Okay, so this week was pretty easy, but it went by too slow. We didn't really learn anything new in class, because we had a test Monday and did study guides the rest of the week. I have to say, though, that I forgot how to do a lot of the problems and stuff from previous chapters. So I guess I'll describe how to do something from an old chapter.
LAW OF COSINES
(opp leg)^2 = (adj leg)^2 + (other adj leg)^2 - 2(adj leg)(adj leg) cos (angle b/w)
You can use an angle to orient yourself like SOHCAHTOA
Example:
a is 5 cm
b is 6 cm
angle C is 36 degrees
find c
c^2 = 5^2 + 6^2 - 2(5)(6) cos(36 degrees)
c = sqrt of ((6^2 + 5^2 - 2(5)(6) cos 36 degrees))
c = 3.53
TRIG INVERSES
you have to find 2 angles usually with trig inverses.
first you need to determine which quadrants your angles are going to be in
(find out where that function is in its sign--ex: sine would be positive in the upper two quadrants, I and II)
then you plug in the positive inverse value into your calculator to get a reference angle, if there is not one given already
quadrant rules:
I to II---make it negative and add 180 degrees
I to III---add 180 degrees
I to IV---make it negative and add 360 degrees
II to IV---add 180 degrees
Example:
cos^-1 (-1/2)
Cosine is negative in the II and III quadrants
Your reference angle is given as 60 degrees.
I to II---make 60 negative and add 180 degrees = 120 degrees
I to III---add 180 degrees to 60 degrees = 240 degrees
Now, the only thing I'm really having problems with are the conics. I always get confused with those. Especially the hyperbolas. Could somebody care to please explain a hyperbola, and maybe do a practice problem? Thanks.
Saturday, December 12, 2009
REFLECTION #17
Well to start off, we didn't learn anything new in this class this week. We've just been doing our study guides to review for our exam. And I have to say, there's a lot of things that kinda forgot how to do from earlier chapers. ha. But anyway, I'm just going to go over a few things we learned early in the year to freshen everyone's memory up because I know I had to look back in my notes to see how to do these things.
The first thing I will go over is completing the square. Completing the square is one of the three ways to solve a quadratic. The other ways are using simple algebra and plugging the numbers into the quadratic formula. The best time to use completing the square when simple algebra does not work is when the linear term is even.
So here is an example:
Ex. 1.) x^2 - 4x = 9
*The first thing you want to do is set your problem up like this:
x^2 - 4x = 9 (by leaving space between the 4x and the equal sign)
*The next thing you do is take your "b" term which is -4 and divide it by 2 and square it
(-4/2)^2 and you get 4
*Now you take that 4 and add it to both sides of the problem like this:
x^2 - 4x + 4 = 9 + 4
*Then you simplify that and get>> (x-2)^2 = square root of 13
*Then take the square root of both sides and get>> x - 2 = +/- square root of 13
*Then add 2 over to the other side and get >> x = 2 +/- square root of 13
**Your final answer in point form is (2 + square root of 13, 0) (2 - square root of 13, 0)
The next thing I will go over is the quadratic form. This is used to solve anything bigger than a quadratic.
Here's an example:
Ex. 2.) x^4 - 7x^2 - 8
*The first thing you do is take "g" and make it equal to x^2. soo g = x^2
*Then you take away the x^2 from every x term and replace the x with "g" and get this:
g^2 - 7g - 8
*Now you solve it like a regular quadratic and you get that g = 8 and g = -1
*Then you take x^2 and set it equal to both 8 and -1 like this:
x^2 = 8 which gives you.... x = +/- 2 square root of 2
and
x^2 = -1 which gives you.... x = +/- i
**Your final answers in point form are (i,0) (-i,0) (2 square root of 2, 0) (-2 square root of 2,0)
Now the last thing I'm going to go over is how to find the inverse of a function and prove that it is an inverse. Keep in mind to graph the function first and see if it passes the horizontal line test. If the line only touches the graph at one point, then it passes. If the line touches the graph at more than one point then it fails the horizontal line test and the function does not have an inverse. And to prove that the inverse is actually an inverse you have to use two functions shown below. And if you get "x" after solving both functions then that means that the inverse is correct.
Here's an example:
Ex. 3.) Find the inverse of y = square root of x-1
*First, if you don't know what the graph of this looks like, then you type it in your calculator and see if it passes the horizontal line test....And it does, so that means that it does have an inverse.
*Next you find the inverse of the equation by first switching the x and y like this:
x = square root of y - 1 Then you solve for y.
*after you solve for y you get that the inverse is y = x^2 + 1
*Now to prove that this is an inverse you have to use these two functions:
1.) F(F^-1(x)) and 2.) F^-1(F(x))
*Let's use the first one first. What this function is telling you to do is plug the inverse equation into the original equation every time you see an x like this:
F(x^2 + 1) =
y = square root of x^2 + 1 - 1
simplifying that you get the square root of x^2 which is x.
*Now use the second function. This function is telling you to plug the original equation into the inverse equation every time you see an x like this:
F^-1(square root of x-1) =
y = (square root of x - 1)^2 + 1
*the square root and ^2 cancel out and you're left with x-1+1 which gives you x.
**So your inverse >> y = x^2 + 1 is an inverse.
*Now for what I don't understand. If anyone can help me with the Ch. 9 "wordproblem" test that would be great because I'm stuck on numbers 2, 4, 7, and 8.
(:
The first thing I will go over is completing the square. Completing the square is one of the three ways to solve a quadratic. The other ways are using simple algebra and plugging the numbers into the quadratic formula. The best time to use completing the square when simple algebra does not work is when the linear term is even.
So here is an example:
Ex. 1.) x^2 - 4x = 9
*The first thing you want to do is set your problem up like this:
x^2 - 4x = 9 (by leaving space between the 4x and the equal sign)
*The next thing you do is take your "b" term which is -4 and divide it by 2 and square it
(-4/2)^2 and you get 4
*Now you take that 4 and add it to both sides of the problem like this:
x^2 - 4x + 4 = 9 + 4
*Then you simplify that and get>> (x-2)^2 = square root of 13
*Then take the square root of both sides and get>> x - 2 = +/- square root of 13
*Then add 2 over to the other side and get >> x = 2 +/- square root of 13
**Your final answer in point form is (2 + square root of 13, 0) (2 - square root of 13, 0)
The next thing I will go over is the quadratic form. This is used to solve anything bigger than a quadratic.
Here's an example:
Ex. 2.) x^4 - 7x^2 - 8
*The first thing you do is take "g" and make it equal to x^2. soo g = x^2
*Then you take away the x^2 from every x term and replace the x with "g" and get this:
g^2 - 7g - 8
*Now you solve it like a regular quadratic and you get that g = 8 and g = -1
*Then you take x^2 and set it equal to both 8 and -1 like this:
x^2 = 8 which gives you.... x = +/- 2 square root of 2
and
x^2 = -1 which gives you.... x = +/- i
**Your final answers in point form are (i,0) (-i,0) (2 square root of 2, 0) (-2 square root of 2,0)
Now the last thing I'm going to go over is how to find the inverse of a function and prove that it is an inverse. Keep in mind to graph the function first and see if it passes the horizontal line test. If the line only touches the graph at one point, then it passes. If the line touches the graph at more than one point then it fails the horizontal line test and the function does not have an inverse. And to prove that the inverse is actually an inverse you have to use two functions shown below. And if you get "x" after solving both functions then that means that the inverse is correct.
Here's an example:
Ex. 3.) Find the inverse of y = square root of x-1
*First, if you don't know what the graph of this looks like, then you type it in your calculator and see if it passes the horizontal line test....And it does, so that means that it does have an inverse.
*Next you find the inverse of the equation by first switching the x and y like this:
x = square root of y - 1 Then you solve for y.
*after you solve for y you get that the inverse is y = x^2 + 1
*Now to prove that this is an inverse you have to use these two functions:
1.) F(F^-1(x)) and 2.) F^-1(F(x))
*Let's use the first one first. What this function is telling you to do is plug the inverse equation into the original equation every time you see an x like this:
F(x^2 + 1) =
y = square root of x^2 + 1 - 1
simplifying that you get the square root of x^2 which is x.
*Now use the second function. This function is telling you to plug the original equation into the inverse equation every time you see an x like this:
F^-1(square root of x-1) =
y = (square root of x - 1)^2 + 1
*the square root and ^2 cancel out and you're left with x-1+1 which gives you x.
**So your inverse >> y = x^2 + 1 is an inverse.
*Now for what I don't understand. If anyone can help me with the Ch. 9 "wordproblem" test that would be great because I'm stuck on numbers 2, 4, 7, and 8.
(:
Reflection
This week we did lots of exam review and we tested our bridges, which turned out our bridge wasn't as strong as we would have like it to be. While doing my review packets i decided to give examples on here from chapter 7. I found the easiest thing with chapter 7 was the conversion from degrees to radians and radians to degrees. Also, converting degrees to degrees, minutes, seconds was simple. The trig chart i have basically memorized by now, but overall pretty simple to comprehend. I'll give the unit circle then a few conversions for examples.
EXAMPLES:
90 degrees, (0,1), pi/2
360 degrees, (1,0), 2pi
180 degrees, (-1,0), pi
270 degrees, (0,-1), 3pi/2
sin = y/r
cos = x/r
tan = y/x
csc = r/y
sec = r/x
cot = x/y
1.) sin 2pi = y/r = 0/1 = 0
2.) cos 180 degrees = x/r = -1/1 = -1
3.) sec 3pi/2 = r/x = 1/0 = undefined
So if anyone wants to help me with the chapter 9 review packet that would be great b/c its the one i'm having the most trouble with. Thanks :)
EXAMPLES:
90 degrees, (0,1), pi/2
360 degrees, (1,0), 2pi
180 degrees, (-1,0), pi
270 degrees, (0,-1), 3pi/2
sin = y/r
cos = x/r
tan = y/x
csc = r/y
sec = r/x
cot = x/y
1.) sin 2pi = y/r = 0/1 = 0
2.) cos 180 degrees = x/r = -1/1 = -1
3.) sec 3pi/2 = r/x = 1/0 = undefined
So if anyone wants to help me with the chapter 9 review packet that would be great b/c its the one i'm having the most trouble with. Thanks :)
Thursday, December 10, 2009
makeup 4
COMPLETING THE SQUARE
1.move # right
2.divide by leading coefficient
3.divide linear term by 2 and square
4.add to both sides
5.factor left
6.solve for x
TO SOLVE QUADRATICS
simple algebra
no linear term
FACTORING
linear is linear is
odd even
quadratic complete the
formula square
GRAPHING PARABOLAS
discriminate-tells how many intercepts graph has
b^2 - 4ac
positive=2x intercepts
negative=no x intercepts
0=1x intercepts
AXIS OF SYMMETRY
x=-b/2a
VERTEX
(-b/2a, f(-b/2a)
SOLVING THE INTERSECTION
solve for y
set=
solve for x
plug back in
1.move # right
2.divide by leading coefficient
3.divide linear term by 2 and square
4.add to both sides
5.factor left
6.solve for x
TO SOLVE QUADRATICS
simple algebra
no linear term
FACTORING
linear is linear is
odd even
quadratic complete the
formula square
GRAPHING PARABOLAS
discriminate-tells how many intercepts graph has
b^2 - 4ac
positive=2x intercepts
negative=no x intercepts
0=1x intercepts
AXIS OF SYMMETRY
x=-b/2a
VERTEX
(-b/2a, f(-b/2a)
SOLVING THE INTERSECTION
solve for y
set=
solve for x
plug back in
makeup 3
FUNCTIONS
1.(f+g)(x)=f(x)+g(x)
2.(f-g)(x)=f(x)-g(x)
3.(f.g)(x)=f(x).g(x)
4.(f/g)(x)=f(x)/g(x)
5.(f0g)(x)=f(g(x))=>composite
6.to reflect on x axis, plug in (-y)
7.to reflect on y axis, plug in (-x)
8.to reflect on y=x find the inverse
a.) switch x and y
b.) solve for y
9.to reflect on origin, 6 and 7
10.if symmetric, you will have same before reflecting
ex: f(x)=x^2 + 1 g(x)=x-3
a.)(f+g)(x)=x^2 +1 + x -3 = x^2+x+2
b.)(f-g)(0)=x^2 +1-(x-3)=4
c.)(f0g)(x)=(x-3)^2 +1=x^2-6x+10
DOMAIN AND RANGE 00=infinity
1.) for all polynomials (-00,00)
Range of all odds (-00,00)
Range of Quadratics (vertex, 00)
2.) fractions
set bottom = 0
solve for x
set up intervals
3.)AV
domain (-00,00)
range(shift, 00)up
(-00, shift)down
4.) square roots
set inside=0
set up # line
try values on either side
eliminate negatives
set up intervals
|-x^2 is a semicircle
1.(f+g)(x)=f(x)+g(x)
2.(f-g)(x)=f(x)-g(x)
3.(f.g)(x)=f(x).g(x)
4.(f/g)(x)=f(x)/g(x)
5.(f0g)(x)=f(g(x))=>composite
6.to reflect on x axis, plug in (-y)
7.to reflect on y axis, plug in (-x)
8.to reflect on y=x find the inverse
a.) switch x and y
b.) solve for y
9.to reflect on origin, 6 and 7
10.if symmetric, you will have same before reflecting
ex: f(x)=x^2 + 1 g(x)=x-3
a.)(f+g)(x)=x^2 +1 + x -3 = x^2+x+2
b.)(f-g)(0)=x^2 +1-(x-3)=4
c.)(f0g)(x)=(x-3)^2 +1=x^2-6x+10
DOMAIN AND RANGE 00=infinity
1.) for all polynomials (-00,00)
Range of all odds (-00,00)
Range of Quadratics (vertex, 00)
2.) fractions
set bottom = 0
solve for x
set up intervals
3.)AV
domain (-00,00)
range(shift, 00)up
(-00, shift)down
4.) square roots
set inside=0
set up # line
try values on either side
eliminate negatives
set up intervals
|-x^2 is a semicircle
makeup 2
Right Triangles
SOH CAH TOA
SOH=sin opp/hyp
CAH=cos adj/hyp
TOA=tan opp/adj
A=90* B=28* b=? c=? a=100
sin28*=b/100
b=100sin28*
b=47
LAW OF SINES
sinA/a = sinB/b = sinC/c
LAW OF COSINES
(opp leg)^2 = (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg)cos(angle between)
ex: x= |6^2 + 5^2 -2(5)(6) cos 36*
x=3.530
SOH CAH TOA
SOH=sin opp/hyp
CAH=cos adj/hyp
TOA=tan opp/adj
A=90* B=28* b=? c=? a=100
sin28*=b/100
b=100sin28*
b=47
LAW OF SINES
sinA/a = sinB/b = sinC/c
LAW OF COSINES
(opp leg)^2 = (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg)cos(angle between)
ex: x= |6^2 + 5^2 -2(5)(6) cos 36*
x=3.530
makeup 1
Trig functions
sin = y/r
cos = x/r
tan = y/x
cot = x/y
sec = r/x
csc = r/y
Trig inverses
1. used to solve for an angle(s)
2. cannot divide by a trig function
3. finds 2 angles, except sin^-1(1), sin^-1(-1)...
4. Steps:
a.)determine the quadratic of the 2 angles
b.)use trig chart to find reference angle (use calculator if not on chart)
c.)Find the 2 angles
5. moving O=theta *=degrees
I > IV make O -ve
I > III add 180*
I > II -O + 180*
II > IV add 180*
6. angles MUST be positive for final answers
sin = y/r
cos = x/r
tan = y/x
cot = x/y
sec = r/x
csc = r/y
Trig inverses
1. used to solve for an angle(s)
2. cannot divide by a trig function
3. finds 2 angles, except sin^-1(1), sin^-1(-1)...
4. Steps:
a.)determine the quadratic of the 2 angles
b.)use trig chart to find reference angle (use calculator if not on chart)
c.)Find the 2 angles
5. moving O=theta *=degrees
I > IV make O -ve
I > III add 180*
I > II -O + 180*
II > IV add 180*
6. angles MUST be positive for final answers
makeup#5
the area of a triangle
the area K of /\ABC is given by:
K = 1/2ab sin C = 1/2bc sin A = 1/2 ac sin B
K = 1/2(one side)*(another side)*(sine of included angle)
2 sides of a triangle have lengths 7cm and 4cm. the angle between the sides measures 73 degrees. find the area.
K = 1/2*7*4*sin73 = 13.4
area = 13.4 cm^2
the area of /\PQR is 15 if p = 5 and q = 10, find all possible measures of angle R
K = 1/2pq sin R
15 = 1/2*5*10*sin R = 25 sin R
R = 15/25 = 0.6
angle R = sin^-1 0.6 = 36.9 degrees OR angle R = 180 - 36.9 = 143.1 degrees
the area K of /\ABC is given by:
K = 1/2ab sin C = 1/2bc sin A = 1/2 ac sin B
K = 1/2(one side)*(another side)*(sine of included angle)
2 sides of a triangle have lengths 7cm and 4cm. the angle between the sides measures 73 degrees. find the area.
K = 1/2*7*4*sin73 = 13.4
area = 13.4 cm^2
the area of /\PQR is 15 if p = 5 and q = 10, find all possible measures of angle R
K = 1/2pq sin R
15 = 1/2*5*10*sin R = 25 sin R
R = 15/25 = 0.6
angle R = sin^-1 0.6 = 36.9 degrees OR angle R = 180 - 36.9 = 143.1 degrees
makeup#4
solving right triangles
sin theta/1 = opposite/hypotenuse cos theta/1 = adjacent/hypotenuse
tan theta = sin theta/cos theta = opposite/adjacent
csc theta = hypotenuse/opposite sec theta = hypotenuse/adjacent
cot theta = adjacent/opposite
for right triangle ABC find the values of b to 3 significant digits
angle = 28 degrees a = 40
to find the value of b, use either tan 28 degrees or cot 28 degrees
tan 28 = opposite/adjacent = 40/b cot 28 = adjacent/opposite
b = 40/tan 28 = 75.2 b = 40cot 28 = 75.2
sin theta/1 = opposite/hypotenuse cos theta/1 = adjacent/hypotenuse
tan theta = sin theta/cos theta = opposite/adjacent
csc theta = hypotenuse/opposite sec theta = hypotenuse/adjacent
cot theta = adjacent/opposite
for right triangle ABC find the values of b to 3 significant digits
angle = 28 degrees a = 40
to find the value of b, use either tan 28 degrees or cot 28 degrees
tan 28 = opposite/adjacent = 40/b cot 28 = adjacent/opposite
b = 40/tan 28 = 75.2 b = 40cot 28 = 75.2
makeup #3
trig functions
1 radian=180/pi degrees = 57.2958 degrees
1 degree=pi/180 radians = 0.0174533 radians
convert 196 degrees to radians
196 degrees = 196 x pi/180 = 3.42 radians
convert 1.35 radians to decimal degrees
1.35 radians = 1.35x180/pi = 77.3 degrees = 77 degrees 20 minutes
coterminal angles
pi/4 find 2 angles, 1 positive 1 negative
positvie angle: pi/4 + 2pi = 9pi/4
negative angle: pi/4 - 2pi = -7pi/4
1 radian=180/pi degrees = 57.2958 degrees
1 degree=pi/180 radians = 0.0174533 radians
convert 196 degrees to radians
196 degrees = 196 x pi/180 = 3.42 radians
convert 1.35 radians to decimal degrees
1.35 radians = 1.35x180/pi = 77.3 degrees = 77 degrees 20 minutes
coterminal angles
pi/4 find 2 angles, 1 positive 1 negative
positvie angle: pi/4 + 2pi = 9pi/4
negative angle: pi/4 - 2pi = -7pi/4
makeup #2
the inclination of a line is the angle alpha, where 0
to the nearest degree, find the inclination of the line 2x+5y=15
y=-2/5x+3
slope=-2/5=tan alpha
alpha=tan^-1(-2/50=21.8 degrees (reference angle)
since tan alpha is negative and alpha is a positive angle, 90 degrees < alpha < 180 degrees
the inclination is 180-21.8=158.2 degrees
to the nearest degree, find the inclination of the line 2x+5y=15
y=-2/5x+3
slope=-2/5=tan alpha
alpha=tan^-1(-2/50=21.8 degrees (reference angle)
since tan alpha is negative and alpha is a positive angle, 90 degrees < alpha < 180 degrees
the inclination is 180-21.8=158.2 degrees
makeup #1
in chapter 8, we learned about simple trigonometric functions
to the nearest 10th of a degree, solve 3cos (-)+9=7 for 0 degrees
3cos (-)+9=7
3cos (-)=-2
cos (-)=2/3 cos^-1 (2/3)=48.2 degrees
(-)=180-48.2=131.8 degrees
(-)=180+48.2=228.2 degrees
these are simple and i understood them almost perfectly. but what i dont get are identies from 8-4... they confuse me and i dont know when to use them or not
to the nearest 10th of a degree, solve 3cos (-)+9=7 for 0 degrees
3cos (-)+9=7
3cos (-)=-2
cos (-)=2/3 cos^-1 (2/3)=48.2 degrees
(-)=180-48.2=131.8 degrees
(-)=180+48.2=228.2 degrees
these are simple and i understood them almost perfectly. but what i dont get are identies from 8-4... they confuse me and i dont know when to use them or not
academic 5/5.
THE TRIG FUNCTIONSSSSS!!!!!
sin= y/r
cos= x/r
tan= y/x
cot= x/y
sec= r/x
csc= r/y
EX: Find all 6 Trig functions for (-3,4)
First you have to find your radius.
r= square root of (-3)^2 + (4)^2
r= square root of 9 + 6
r= square root of 25
r= 5
Now find all of the functions.
sin= 4/5
cos= -3/5
tan= 4/-3
cot= -3/4
sec= 5/-3
csc= 5/4
The trig chart is easy if you study.
0 degrees: sin 0 = 0
30 degrees: sin pi/6 = 1/2
45 degrees: sin pi/4 = square root of 2/ 2
60 degrees: sin pi/3 = square root of 3/ 2
90 degrees: sin pi/2 = 1
Now for cos, the chart just flips, so cos pi/2 = 0 and cos 0 = 1
tan 0 = 0
tan pi/6 = square root of 3/ 2
tan pi/4 = 1
tan pi/3 = square root of 3
tan pi/2 = undefined
Now for cot, the chart just flips, so cot 0 = undefined and cot pi/2 = 0.
sec 0 = 1
sec pi/6 = 2 square root of 3/ 3
sec pi/4 = square root of 2
sec pi/3 = 2
sec pi/2 = undefined
Now for csc, the chart just flips, so csc 0 = undefined and csc pi/2 = 1.
remember that and study for 15 minuets a day and your set to go!
When you find a coterminal angle, you just add or subtract 360 degrees until you get your answer.
If you are asked to find the negative coterminal angle of 2234,
you just keep subtracting 360 degrees until you get a negative number.
sin= y/r
cos= x/r
tan= y/x
cot= x/y
sec= r/x
csc= r/y
EX: Find all 6 Trig functions for (-3,4)
First you have to find your radius.
r= square root of (-3)^2 + (4)^2
r= square root of 9 + 6
r= square root of 25
r= 5
Now find all of the functions.
sin= 4/5
cos= -3/5
tan= 4/-3
cot= -3/4
sec= 5/-3
csc= 5/4
The trig chart is easy if you study.
0 degrees: sin 0 = 0
30 degrees: sin pi/6 = 1/2
45 degrees: sin pi/4 = square root of 2/ 2
60 degrees: sin pi/3 = square root of 3/ 2
90 degrees: sin pi/2 = 1
Now for cos, the chart just flips, so cos pi/2 = 0 and cos 0 = 1
tan 0 = 0
tan pi/6 = square root of 3/ 2
tan pi/4 = 1
tan pi/3 = square root of 3
tan pi/2 = undefined
Now for cot, the chart just flips, so cot 0 = undefined and cot pi/2 = 0.
sec 0 = 1
sec pi/6 = 2 square root of 3/ 3
sec pi/4 = square root of 2
sec pi/3 = 2
sec pi/2 = undefined
Now for csc, the chart just flips, so csc 0 = undefined and csc pi/2 = 1.
remember that and study for 15 minuets a day and your set to go!
When you find a coterminal angle, you just add or subtract 360 degrees until you get your answer.
If you are asked to find the negative coterminal angle of 2234,
you just keep subtracting 360 degrees until you get a negative number.
academic. 4/5
Inequalities and Finding the Domain and Range.
Chapter 8. review.
solveing an inequality, just treat it as a regular equation like in algebra.. **but dont forget when you divide by a -ve you must switch the sign.
here is an example:
-2x + 5 lt 7
-2x lt 2
x gt -1 (You switch the sign because you have to divide by -2)
for absolute inequalities its a little different. You have to come out with two answers because they can be +ve or -ve.
If the first sign (the one in the problem) is > greater than or = 2, or just greater than, it is called: or problem.
If the first sign is < less than or = 2, or just less than, it is called an and problem.
First, you get the absolute value by itself
Then, you set up two different problems out of the original.
One problem is the original.
The second problem:
you switch the sign of the value,
and switch the sign of the constant on the other side.
Example:
2x + 3 lt 6 (the absolute value is already by itself, so form two equations)
2x + 3 lt 6 and 2x + 3 gt -6
2x lt 3 and 2x gt -9
x lt 3/2 and x gt -9/2
Now before you make that your final answer, you have to check. The answer that works is the one your circle for your answer.
3/2 does not work because if you plug it in, 6 lt 6.
-9/2 does work because -6 lt 6.
So, your answer would simply be -9/2
Chapter 8. review.
solveing an inequality, just treat it as a regular equation like in algebra.. **but dont forget when you divide by a -ve you must switch the sign.
here is an example:
-2x + 5 lt 7
-2x lt 2
x gt -1 (You switch the sign because you have to divide by -2)
for absolute inequalities its a little different. You have to come out with two answers because they can be +ve or -ve.
If the first sign (the one in the problem) is > greater than or = 2, or just greater than, it is called: or problem.
If the first sign is < less than or = 2, or just less than, it is called an and problem.
First, you get the absolute value by itself
Then, you set up two different problems out of the original.
One problem is the original.
The second problem:
you switch the sign of the value,
and switch the sign of the constant on the other side.
Example:
2x + 3 lt 6 (the absolute value is already by itself, so form two equations)
2x + 3 lt 6 and 2x + 3 gt -6
2x lt 3 and 2x gt -9
x lt 3/2 and x gt -9/2
Now before you make that your final answer, you have to check. The answer that works is the one your circle for your answer.
3/2 does not work because if you plug it in, 6 lt 6.
-9/2 does work because -6 lt 6.
So, your answer would simply be -9/2
academic 3/5
Chapter 9. Section 3.
LAW OF SINES.
sinA/a=sinB/b=sinC/c
*the law of sines is used when you know pairs in non right triangles.
*also use the law of sines when you are setting up a proportion.
A civil engineer want to determine the distances from point A and B to an inaccessible point C. From the direct measurement the engineer know AB= 25 m
Sin50(degree0/25 = sin 20 (degree)/ b
SOLVE for b:
Bsin20=25sin50
b=25sin50/sin20
Approx. = 11.162m
Example:
Sin A/123 = sin A/16
16sin115/123= 1233m
Sin A= 16 sin 115/123
A=sin-1( (16sin 115)/ 123)
A approx. = 6.771 degrees
Can someone explain Identities and Equations to me, I get the basic ones its when you have to prove or have things like sin^2X-cos^2X / cos^2 X.
LAW OF SINES.
sinA/a=sinB/b=sinC/c
*the law of sines is used when you know pairs in non right triangles.
*also use the law of sines when you are setting up a proportion.
A civil engineer want to determine the distances from point A and B to an inaccessible point C. From the direct measurement the engineer know AB= 25 m
Sin50(degree0/25 = sin 20 (degree)/ b
SOLVE for b:
Bsin20=25sin50
b=25sin50/sin20
Approx. = 11.162m
Example:
Sin A/123 = sin A/16
16sin115/123= 1233m
Sin A= 16 sin 115/123
A=sin-1( (16sin 115)/ 123)
A approx. = 6.771 degrees
Can someone explain Identities and Equations to me, I get the basic ones its when you have to prove or have things like sin^2X-cos^2X / cos^2 X.
makeup blog #5
how to graph a trig function..........
EX: y=2sin(3X+pie)-4
1. amp = 2
2. period = (2pie/3) <---leave pie out and (p/4) = (1/6) and add to each
1. 0
2. pie/6
3. pie/3
4. pie/2
5. 2pie/3
3. phase shift = -pie
1. 0 -pie = -pie
2. pie/6 -pie = -5pie/6
3. pie/3 - pie = -2pie/3
4. pie/2 - pie = -pie/2
5. 2pie/3 - pie = pie/3
y= 2sin(3X+pie)-4
....then you simply draw graph
4. vertical shift = 4
-one thing i didnt understand was identites and equations..
i think my biggest issue is actually sitting down and memorizing the entire identity chart
academic 2/5
Chapter 9. Trigonometry section 2.
Area of a non right triangle:
A= 1/2 (leg)(leg)sin(angle b/w)
when two sides of a triangle have lengths 7cm the angle between the sides meansure 73degrees. Find the area: ** it may help if you sketch the triangle i just described.
A= 1/2 (7)(4) sin (73)
approx = 13.388 cm^2
Remember to always use the approx sign when you write the answer to these problems because it is not exact.
another example:
The area of tri PQR is 15. If p=5 and q=10 find all possible measures. now plug it into your formula and don’t forget all your coterminal angles also!
A= 15
15= 1/2(5)(10)sinR
15=25sinR
sinR=3/5
R=sin-1(3/5)
R= 36.870 degrees, 143.130 degrees.
When you do the coterminal angles you are making the original answer -ve and either adding or subtracting 180 or 360.
Area of a non right triangle:
A= 1/2 (leg)(leg)sin(angle b/w)
when two sides of a triangle have lengths 7cm the angle between the sides meansure 73degrees. Find the area: ** it may help if you sketch the triangle i just described.
A= 1/2 (7)(4) sin (73)
approx = 13.388 cm^2
Remember to always use the approx sign when you write the answer to these problems because it is not exact.
another example:
The area of tri PQR is 15. If p=5 and q=10 find all possible measures. now plug it into your formula and don’t forget all your coterminal angles also!
A= 15
15= 1/2(5)(10)sinR
15=25sinR
sinR=3/5
R=sin-1(3/5)
R= 36.870 degrees, 143.130 degrees.
When you do the coterminal angles you are making the original answer -ve and either adding or subtracting 180 or 360.
makeup blog #4
triangle ABC.
andgle c is 36 and b is 7...find sides a and c
use SOHCAHTOA
lets find c,
set up equation which is...........
tan36=(c/7)
c is app. 5.086
side a
sin36=(5.086/a)
a is app. 8.653
find area = A=(1/2)bh
Area is app. 17.801
makeup blog #3
angle of inclination
10X + 5y = 15
find slope which is = -2
m=tan theata
tan theata=-2
theata=tan(inverse)(2) = 63.435
quadrants are negative in second and forth then 63.435 make it negative and add 360 and get 296.565
tan theata is app. 116.565 and 295.565
makeup blog #2
law of cosines
(opp leg)=(adj leg)^2 + (other adj leg)-2(adj leg) (adj leg) cos (angle between)
it is used when solving for non-right triangles
SOHCAHTOA is also the law of sines
law of sines
EX: angle B=30 degrees, angle A=135 degrees, and side b=4
...find C and a and c.
find your other angle, so add 30 to 135 together and get 165, then you subreact that from 180 and get 15 so angle C is 15 degrees.
next, find another side...since you have pairs, you ca use that in the law of sines
your equation would be sin(30/4)=sin(135/a)
cross multiple and get asin(30)= 4sin15
........which means that c is app. 2.071
I STILL DONT GET ALL THE WORD PROBLEMS IN THIS SECTION..
makeup blog #1
in chapter 9 what i understand most is trig functions
EX: cos (inverse) (-1/2)
cos is dealing with the x axis and since (-1/2) is negative you're looking for where cos is negative which are the second and third quadrants. (1/2) is on your trig chart and is 60 degrees. to find your other two angles you first have to make move from the first to second quadrant...you make 60 negative and add 180 which gives you 120 degrees. then you find your second, you have to move from the first quad. to the third. you simply add 180 to 60 and get 240 degrees.
.........im still a little uncertain about the right triangles, any takers?
Wednesday, December 9, 2009
academic 1/5
Chapter 7. Trigonometry Section 1.
Alright an easy way to convert angles to radians is by using the formula pi/180 The only thing you have to remember is that when you find the answer on your calculator instead of typing in pi use 1. using pi will give you a huge answer.
example:
Ydegrees x pi/180 plugged in as Ydegrees x 1/180
250degrees x pi/180= 225/180pi= 5/4pi
pretty simple right? here is your hint, well more of this is what has to be done- you must use the degree symbol at all times for answers in degrees. Sense your working in radians and degrees their is no other way to tell what function your working in besides the degree symbol. Also remember to ALWAYS keep your calculator in the degree function if not it will give you all the wrong answer and keep checking the setting after you work a few problems.
Another easy thing to do is to convert degrees, minuets, and seconds back to degrees:
25degrees 20’ 6”
25 + 20/60 + 6/3600= 25.335 degrees
All your doing is keeping your whole number, diving 20 by 60 because you used 60 to get there in the first place and then dividing 6 by 3600 for the same reason. This is all working the equation backwards. So to find the degree’s minutes and seconds your multiply each number by 60 and 3600.
15 degrees 24’ 15”
15+ 24/60 + 15/3600 = 15.404 degrees
Alright an easy way to convert angles to radians is by using the formula pi/180 The only thing you have to remember is that when you find the answer on your calculator instead of typing in pi use 1. using pi will give you a huge answer.
example:
Ydegrees x pi/180 plugged in as Ydegrees x 1/180
250degrees x pi/180= 225/180pi= 5/4pi
pretty simple right? here is your hint, well more of this is what has to be done- you must use the degree symbol at all times for answers in degrees. Sense your working in radians and degrees their is no other way to tell what function your working in besides the degree symbol. Also remember to ALWAYS keep your calculator in the degree function if not it will give you all the wrong answer and keep checking the setting after you work a few problems.
Another easy thing to do is to convert degrees, minuets, and seconds back to degrees:
25degrees 20’ 6”
25 + 20/60 + 6/3600= 25.335 degrees
All your doing is keeping your whole number, diving 20 by 60 because you used 60 to get there in the first place and then dividing 6 by 3600 for the same reason. This is all working the equation backwards. So to find the degree’s minutes and seconds your multiply each number by 60 and 3600.
15 degrees 24’ 15”
15+ 24/60 + 15/3600 = 15.404 degrees
Chapter 6,7,and 9 review
This blog is dealing with the review of chapter 6,7,and 9 on the exam review. Going over theses rules should refresh you of the chapters and help alot on the exams. In chapter 6 we did conics. Equation of a circle in standard form is(x-h)^2+(y-k)=r^2. If it is not in standard form you must complete the square to put in the standard form. Given the center and a point you can use the distance formula to find the radius. There are four steps to find the intersection of a line and a circle. The steps are the following:solve the linear equation for g, substitute a circle equation, solve for x, plug x value to get the y value. Note if your x value is imaginary then there is no point of intersection. To sketch the shape of the circle you must shape, center, find the major and minor, other int, focus, asymototes, vertex, ND SKETCH. no. How to know the shape of a graph if not in standard for, well look at your notes and you shall see 3 steps. If-ve then it is a circle or an ellipse to be a circle B=)0 then it is a parobola, is positive then hyperbola. Trignometry-angles measured in degrees minutes, and seconds. To find minutes multiply what is behind deimal by 60. To seconds multiply what is behind the deciaml by 60 and divide by 3600 to get decimal. Angles are measured in degrees and radians . Always use exact answer do not plug pie. To find the cotermianl angle add or subtract 360. Hints- must use symbol if in degrees or it is wrong. If no degree symbol it is assumed that you are in radians. S=Rtheta, k=1/2r2theta. Here some more notes, sin=y'r, cos=x/r tan=y/x, csc sec and cot or the same except flipped or reversed.. Study your Trig chART.. We did alot of trig inverses too, so study that. We learned the law of sines,and the law of cosines!
Chapter 5 and 6 exam review
This information given should help with the second quarter exam. All examples shown should help with the chapter 5 and 6 study guide, and most importantly it should help on the midterm exam. We learned how to deal with exponents. When you are multiplying numbers you add the exponent, when dividing exponents you subtract the exponents, when you have to different variable you combine the exponents, when you divide to different variables you get to different exponents, multiplying to exponents you getbxy. To solve for an exponent write as the same base, set exponents equal, and solve for x. Sometimes you will have to sandwhich a problem. WE went over logs, logs is just another pig latin term for math. Here are some log properties. Logmn=logbm+logbn, logbmk=klogbm. logbbx=k, and blog10k=k. You can expand logs, condense logs, and express y in terms of x. Change of a Base, used when a log cant be solved, used to solve for x as a variable, and used to change the base of the log. Here are the steps, write as an expotiential, take the log of both sides, move exponent to the front, sove for variable, write as a fraction of whole number if possiable. If not possiable leave in log form. We did some expotiential functions. We learned how to do interest rate and solved a few problems doing that. And that all we did for chapters 5 and 6.
Makeup Reflection #8
I understand all of the problems when you have to find the inverses, this is the easiest. All these problems you have to find two angles on the coordinate plane.
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Here are some helpful hints that you can use on your homework or your tests.
**secx=3
x=sec^-1(3)
x=cos^-1(1/3)
^^they are the same because they are related. All you have to do is switch the sec to cos, so you flip the three and the one, and 3 becomes 1/3.
**cscx=6
x=csc^-1(6)
x=sin^-1(1/6)
^^this is the same thing from the beginning, you just switch the 6 and it becomes 1/6
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EX:
sinx= -0.7
x=sin^-1(0.7)
x=224.42degrees, 315.57degrees
sin is negative the third and fourth quadrant, so you are looking for the angles in both of those problems.
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EX:
tanx=1.2
x=tan^-1(1.2)
x=50.19degrees, 230.2degrees
tan is positive in the first and the third quadrant, so you need to find all of the angles.
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EX:
2tanx+1=0 on 0 less than equal to x less than or equal to pi.
tanx= -1/2
x=tan^-1(-1/2)
x= 153.435degrees x pi/180
x= .852pi
tan is negative in the second and fourth quadrant. So you need to find the two angles in both of the quadrants.
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The thing that i did not understand was the identities. I don't understand how you can cross out all of them, but not all of it. If someone can help me out with a problem, please show me an example. THANKSSSS.
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Here are some helpful hints that you can use on your homework or your tests.
**secx=3
x=sec^-1(3)
x=cos^-1(1/3)
^^they are the same because they are related. All you have to do is switch the sec to cos, so you flip the three and the one, and 3 becomes 1/3.
**cscx=6
x=csc^-1(6)
x=sin^-1(1/6)
^^this is the same thing from the beginning, you just switch the 6 and it becomes 1/6
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EX:
sinx= -0.7
x=sin^-1(0.7)
x=224.42degrees, 315.57degrees
sin is negative the third and fourth quadrant, so you are looking for the angles in both of those problems.
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EX:
tanx=1.2
x=tan^-1(1.2)
x=50.19degrees, 230.2degrees
tan is positive in the first and the third quadrant, so you need to find all of the angles.
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EX:
2tanx+1=0 on 0 less than equal to x less than or equal to pi.
tanx= -1/2
x=tan^-1(-1/2)
x= 153.435degrees x pi/180
x= .852pi
tan is negative in the second and fourth quadrant. So you need to find the two angles in both of the quadrants.
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The thing that i did not understand was the identities. I don't understand how you can cross out all of them, but not all of it. If someone can help me out with a problem, please show me an example. THANKSSSS.
Chapter 3 and 4 exam review
This information given should help with the second quarter exam. All examples shown should help with the chapter 3 and 4 study guide, and most importantly it should help on the midterm exam. In chapter 3 we learned how to sketch polonomial factors. Factor completely2.set up a number line and label zeros3.Plug in on either side of your root4.positive is above x axis and negatives or below the x axis5.check in calculater6.max and min is calculated onlyto find the max and min you do a negative b divided by 2a. We also did a few problems with finding the area of a few figures. We learned how to do inequalities in this chapter. Inequalities-change sign when you multiply or divide by a negative. Example problem= [3x-9]>4 x-9>4.We had many homework problems on page 98 #1-24all. We had many notes for domain and range.
1.Domain of all polynomials are infinity. The rand will always go up or down depending if it is positive or negative. When dealing with a fraction you set the bottom equal to zero solve for x and then set up an interval. Absolut valur domain is a infinity number, and the range can sdhift but it is also an infinity number. To do the square root you set a side =0, set up a number line, try values on either side, eliminate anything negative, and then set up intervals. Beware of a negative square root because it has a limited range. Range is from top to bottom and domain is left to right. We also did a bunch of f of x and g of x problems with adding subtracting multiplying and dividing. We learned how to do an inverse. To find an inverse swithch x and y and solve for y. To have an inverse it must pass the horizontal line test.. Here some helpful hints: anything done inside shifts the opposite or x axis, anything done on the outside shifts the y axis, if less than one it is fatter and taller, if greater than 1 it makes it skinny and shrorter.
1.Domain of all polynomials are infinity. The rand will always go up or down depending if it is positive or negative. When dealing with a fraction you set the bottom equal to zero solve for x and then set up an interval. Absolut valur domain is a infinity number, and the range can sdhift but it is also an infinity number. To do the square root you set a side =0, set up a number line, try values on either side, eliminate anything negative, and then set up intervals. Beware of a negative square root because it has a limited range. Range is from top to bottom and domain is left to right. We also did a bunch of f of x and g of x problems with adding subtracting multiplying and dividing. We learned how to do an inverse. To find an inverse swithch x and y and solve for y. To have an inverse it must pass the horizontal line test.. Here some helpful hints: anything done inside shifts the opposite or x axis, anything done on the outside shifts the y axis, if less than one it is fatter and taller, if greater than 1 it makes it skinny and shrorter.
Comment 1 and 2
Comment 1
This is a review for chapter 1 for the exam.
Completing the square there is 6 steps
1.move number to the right
2.divide by leading coefficent
3.divide linear term by 2 and square it
4.add to both sides
5.factor by left
6.solve for x
Comment 2
This is a review to sketch for a polynomial function
1.Factor completely
2.set up a number line and label zeros
3.Plug in on either side of your root
4.positive is above x axis and negatives or below the x axis
5.check in calculater
6.max and min is calculated only
to find the max and min you do a negative b divided by 2a
This is a review for chapter 1 for the exam.
Completing the square there is 6 steps
1.move number to the right
2.divide by leading coefficent
3.divide linear term by 2 and square it
4.add to both sides
5.factor by left
6.solve for x
Comment 2
This is a review to sketch for a polynomial function
1.Factor completely
2.set up a number line and label zeros
3.Plug in on either side of your root
4.positive is above x axis and negatives or below the x axis
5.check in calculater
6.max and min is calculated only
to find the max and min you do a negative b divided by 2a
Makeup Reflection #7
You use law of cosines when you are looking for a side of a triangle. I think that this is the easiest laws to use because it gives you all of the information you need to solve the equation, triagle.
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Law of cosines:
(opp leg)^2 = (adj.leg)^2 + (other adj.leg)^2 - 2(adj.leg)(adj.leg)cos(anlge between)
-Use an angle to orient yourself like SOHCAHTOA.
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EX: x^2= 6^2 + 5^2 - 2(5)(6)cos36degrees
x= sqareroot of 6^2 + 5^2 -2(5)(6) cos36degrees
x= 3.530
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EX: 7^2 = 6^2 + 5^2 - 2(5)(6)cos(alpha)
7^2 - 6^2 - 5^2 = -2(5)(6)cos(alpha)
cos(alpha) = 7^2-6^2-5^2/-2(5)(6)
(alpha)= cos^-1((7^2-6^2-5^2)/(-2(5)(6))) <---plug this all into your calculator
alpha= 78.436degrees.
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I really don't have any problems with these kind of laws, i just could use some more examples. Some of the hard problems, like the ones where they do not give you enough information, i could use a couple of those. But other than that, i pretty much understand this lesson. I also understand the law of sines. They are both pretty easy to me. But if anyone wants to copy me a problem down, go right ahead, i would greatly appreciatie it. THANKSSSSS.
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Law of cosines:
(opp leg)^2 = (adj.leg)^2 + (other adj.leg)^2 - 2(adj.leg)(adj.leg)cos(anlge between)
-Use an angle to orient yourself like SOHCAHTOA.
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EX: x^2= 6^2 + 5^2 - 2(5)(6)cos36degrees
x= sqareroot of 6^2 + 5^2 -2(5)(6) cos36degrees
x= 3.530
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EX: 7^2 = 6^2 + 5^2 - 2(5)(6)cos(alpha)
7^2 - 6^2 - 5^2 = -2(5)(6)cos(alpha)
cos(alpha) = 7^2-6^2-5^2/-2(5)(6)
(alpha)= cos^-1((7^2-6^2-5^2)/(-2(5)(6))) <---plug this all into your calculator
alpha= 78.436degrees.
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I really don't have any problems with these kind of laws, i just could use some more examples. Some of the hard problems, like the ones where they do not give you enough information, i could use a couple of those. But other than that, i pretty much understand this lesson. I also understand the law of sines. They are both pretty easy to me. But if anyone wants to copy me a problem down, go right ahead, i would greatly appreciatie it. THANKSSSSS.
Makeup Reflection #6
Here is an example of the amplitude problem.
Page 328, #6:
Amplitude = total distance/2 = 4/2= 2
y=2sin(pi/2x+___)+1
2pi/B= 4
2pi/B= 2pi/4
pi/2
PERIOD= 2pi/B
y=2sin(pi/2x)+3 <------ find the equation of the graph.
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EX: y=3cos(pix-2)+1
amplitude= 3
period= 2
vertical shift= 1
phase shift= 3
1. 0
2. 1/2
3. 1
4. 3/2
5. 2
FINAL GRAPH
1. 0+2= 2
2. 1/2+2= 5/2
3. 1+2= 3
4. 3/2+2= 7/2
5. 2+2= 4
y=3cos(pix-2)+1
**you have to graph your first graph and your final graph on the same coordinate plane, you just have to use different color.
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The thing that I didn't understand was where you start on the graph, I know that when you have sin, you start on the x-axis. And when you have cos, you start at the maximum y-value. But I do not understand where you start if they do not give you the starting point.
Page 328, #6:
Amplitude = total distance/2 = 4/2= 2
y=2sin(pi/2x+___)+1
2pi/B= 4
2pi/B= 2pi/4
pi/2
PERIOD= 2pi/B
y=2sin(pi/2x)+3 <------ find the equation of the graph.
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EX: y=3cos(pix-2)+1
amplitude= 3
period= 2
vertical shift= 1
phase shift= 3
1. 0
2. 1/2
3. 1
4. 3/2
5. 2
FINAL GRAPH
1. 0+2= 2
2. 1/2+2= 5/2
3. 1+2= 3
4. 3/2+2= 7/2
5. 2+2= 4
y=3cos(pix-2)+1
**you have to graph your first graph and your final graph on the same coordinate plane, you just have to use different color.
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The thing that I didn't understand was where you start on the graph, I know that when you have sin, you start on the x-axis. And when you have cos, you start at the maximum y-value. But I do not understand where you start if they do not give you the starting point.
Makeup Reflection #5
I didn't quite get the identities, but i copied all of them down. From the book and started to memorize the and it helped me out because now i know some of the identities and how to somewhat work the problems.
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IDENTITIES:
Reciprocal Relations
cscx= 1/sinx
secx= 1/cosx
cotx= 1/tanx
Relations with Negatives
sin(-x)= -sinx & cos(-x)= -cosx
csc(-x)= -cscx & sec(-x)= -secx
tan(-x)= -tanx & cot(-x)= -cotx
Pythagorean Relationships
sin^2x+cos^2x=1
1+tan^2x=sec^2x
1+cot^2x=csc^2x
Cofunction Relationships
sinx= cos(90degrees - x) & cosx=sin(90degrees - x)
tanx= cot(90degrees - x) & cotx=tan(90degrees - x)
secx= csc(90degrees - x) & cscx=sec(90degrees - x)
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EX: Simplify secx-sinxtanx
1. Identities
2. Algebra
tanx= sinx/cosx
cotx= cosx/sinx
1/cosx - sinx(sinx/cosx)
1/cosx - sin^2x/cosx
1-sin^2x/cosx
sin^2 + cos^2 = 1
-sin -sin
cos^2x/cosx
^^the cosx cancels out and you are left with cosx.
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I didn't get this at first, but now since i learned all the identities and had more practice with it, i sorta grasped the concept. But if someone would like to give me some examples, go right ahead, THANKSSS
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IDENTITIES:
Reciprocal Relations
cscx= 1/sinx
secx= 1/cosx
cotx= 1/tanx
Relations with Negatives
sin(-x)= -sinx & cos(-x)= -cosx
csc(-x)= -cscx & sec(-x)= -secx
tan(-x)= -tanx & cot(-x)= -cotx
Pythagorean Relationships
sin^2x+cos^2x=1
1+tan^2x=sec^2x
1+cot^2x=csc^2x
Cofunction Relationships
sinx= cos(90degrees - x) & cosx=sin(90degrees - x)
tanx= cot(90degrees - x) & cotx=tan(90degrees - x)
secx= csc(90degrees - x) & cscx=sec(90degrees - x)
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EX: Simplify secx-sinxtanx
1. Identities
2. Algebra
tanx= sinx/cosx
cotx= cosx/sinx
1/cosx - sinx(sinx/cosx)
1/cosx - sin^2x/cosx
1-sin^2x/cosx
sin^2 + cos^2 = 1
-sin -sin
cos^2x/cosx
^^the cosx cancels out and you are left with cosx.
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I didn't get this at first, but now since i learned all the identities and had more practice with it, i sorta grasped the concept. But if someone would like to give me some examples, go right ahead, THANKSSS
Reflection Make Up #2
For this reflection I am going to talk about inverses because, well.......I jus feel like talkin bout inverses.
Some things to know about inverses:
-used to solve for an angle.
-You cannot divide by a trig function.......ever!
-An inverse finds 2 angles except for the exceptions:
inverse sin (1), inverse sine (-1), inverse csc(1), inverse csc (-1)
inverse cos (1), inverse cos (-1), invers sec (1), inverse sec (-1)
And now......the steps to finding an inverse of a trig function.
1) Determine the quadrant of your 2 angles
2) Use trig chart to find reference angle, if not on chart or unit circle use calculator.
3) find the 2 angles.
-To move quadrants:
I--IV make angle -ve and add 360
I--III add 180
I--II make angle negative and add 180
II--IV add 180
Example:
inverse csc (2) = reference angle= 30 degrees.
To find the quadrants you want, look at the number and see if it is +ve or -ve. In this one it is positive, so look for where csc is positive. csc is positive in the quadrants I and II so 30 is one of your angles, to get the other, make it negative and add 180.
-30 + 180 = 150
So your second angle is 150 degrees.
Some things to know about inverses:
-used to solve for an angle.
-You cannot divide by a trig function.......ever!
-An inverse finds 2 angles except for the exceptions:
inverse sin (1), inverse sine (-1), inverse csc(1), inverse csc (-1)
inverse cos (1), inverse cos (-1), invers sec (1), inverse sec (-1)
And now......the steps to finding an inverse of a trig function.
1) Determine the quadrant of your 2 angles
2) Use trig chart to find reference angle, if not on chart or unit circle use calculator.
3) find the 2 angles.
-To move quadrants:
I--IV make angle -ve and add 360
I--III add 180
I--II make angle negative and add 180
II--IV add 180
Example:
inverse csc (2) = reference angle= 30 degrees.
To find the quadrants you want, look at the number and see if it is +ve or -ve. In this one it is positive, so look for where csc is positive. csc is positive in the quadrants I and II so 30 is one of your angles, to get the other, make it negative and add 180.
-30 + 180 = 150
So your second angle is 150 degrees.
Reflection Make Up #3
Once again, in the spirit of reviewing for the mid term exam i'ma go back in time al lil bit to the beginning of trigonometry. Some people asked me today how to find all six trig functions when you are givin just one point on a graph. Here we go.
Say you are given the point (1,-1) and told to find all six trig functions:
-first graph the point
-now draw a line from the point you graphed to (0,0)
-now use the x axis as the last side of the triangle...thats right, triangle. You now have a right triangle with sides 1 and 1.
-so now use the pythagorean theorem to find the hypotenuse (r).
1^2 + 1^2 = c^2
1 + 1 = 2
square root of 2
-now plug in to your formulas for the six trig functions:
sin = y/r
cos = x/r
tan = y/x
csc = r/y
sec = r/x
cot = x/y
-the x axis leg of the triangle is 1, so x = 1
-the y axis leg of the triangle is -1, so y = -1
-the hypotenuse of the triangle is the square root of 2, so r = square root of 2
-so the answers are:
sin = -1/square root of 2 = -square root of 2/2
cos = 1/square root of 2 = square root of 2/2
tan = -1/1 = -1
csc = square root of 2/-1 = -square root of 2
sec = square root of 2/1 = square root of 2
cot = 1/-1 = -1
This is a very simple section of trigonometry, but if you dont know what your trig functions equal you have no chance at this (along with the rest of trig). Also remember to always draw out your triangle first.
Say you are given the point (1,-1) and told to find all six trig functions:
-first graph the point
-now draw a line from the point you graphed to (0,0)
-now use the x axis as the last side of the triangle...thats right, triangle. You now have a right triangle with sides 1 and 1.
-so now use the pythagorean theorem to find the hypotenuse (r).
1^2 + 1^2 = c^2
1 + 1 = 2
square root of 2
-now plug in to your formulas for the six trig functions:
sin = y/r
cos = x/r
tan = y/x
csc = r/y
sec = r/x
cot = x/y
-the x axis leg of the triangle is 1, so x = 1
-the y axis leg of the triangle is -1, so y = -1
-the hypotenuse of the triangle is the square root of 2, so r = square root of 2
-so the answers are:
sin = -1/square root of 2 = -square root of 2/2
cos = 1/square root of 2 = square root of 2/2
tan = -1/1 = -1
csc = square root of 2/-1 = -square root of 2
sec = square root of 2/1 = square root of 2
cot = 1/-1 = -1
This is a very simple section of trigonometry, but if you dont know what your trig functions equal you have no chance at this (along with the rest of trig). Also remember to always draw out your triangle first.
Reflection Make Up 1
One thing I never blogged about yet is the law of sines, and with exams coming back I figured I'd backtrack a little for some review. When we first started learning law of sines, for some reason I found it ridiculously hard, but it got easy once you grasp the process. The law of sines is used when there is a known pair, that is, a know angle and the side opposite of the angle. When you set it up, you set it up just like a proportion:
sinA/a = sinB/b = sinC/c
Here is an example of finding an unknown angle using the law of sines:
Triangle ABC has sides b = 123, a = 16 and angle B = 115 degrees. Find angle A
-now we know we can use law of sines because of the angle B is know and side b is known.
-so now set up the proportion:
sin115/123 = sinA/16
now cross multiply
sinA = 16 sin 115/123 now take an inverse:
A = sin^-1(16 sin 115/123) which equals aproximately 6.771 degrees
-So angle A =6.771 degrees.
Now here is an example of how to find a side:
Triangle ABC has side a = 4 and angles A = 30 degrees and C = 25 degrees. Find b.
-we know we can use law of sines because angle A and side a match up.
-set a your proportion a little bit different:
sin30/4 = sin25/b cross multiply:
b sin 30=4 sin 25 solve for b:
b = 4 sin 25/sin 30
-Side b = 3.381
___________________________________________________
I don't remember how to find the intersection of 2 lines when you are given the equations of the lines.
sinA/a = sinB/b = sinC/c
Here is an example of finding an unknown angle using the law of sines:
Triangle ABC has sides b = 123, a = 16 and angle B = 115 degrees. Find angle A
-now we know we can use law of sines because of the angle B is know and side b is known.
-so now set up the proportion:
sin115/123 = sinA/16
now cross multiply
sinA = 16 sin 115/123 now take an inverse:
A = sin^-1(16 sin 115/123) which equals aproximately 6.771 degrees
-So angle A =6.771 degrees.
Now here is an example of how to find a side:
Triangle ABC has side a = 4 and angles A = 30 degrees and C = 25 degrees. Find b.
-we know we can use law of sines because angle A and side a match up.
-set a your proportion a little bit different:
sin30/4 = sin25/b cross multiply:
b sin 30=4 sin 25 solve for b:
b = 4 sin 25/sin 30
-Side b = 3.381
___________________________________________________
I don't remember how to find the intersection of 2 lines when you are given the equations of the lines.
Makeup Reflection #4
**caution
1. You can not divide trig functions to cancel them
-you can move everything to one side-or-factor out a trig function.
-you can divide by a trig function to create a new one.
_______________________________________________________________________
EX: sinxtanx=3sinx for 0 less than or equal to x less than or equal to 2pi.
sinxtanx-3sinx=0
sinx(tanx-3)<------ factor out a sinx
sinx=0 tanx-3=0
x=sin^-1(0) x=tan^-1(3)
x= 0, pi, 2pi x= 71.565degrees, 251.565degrees
^^you need to use the unit cirlce for this problem because it asks for the inverse of sin of 0, and if you use your unit circle, it is much easier to know it. It conserves time so you don't have to plug it into your caculator.
_______________________________________________________________________
EX: 2sinx=cosx for 0 less than or equal to x less than or equal to 360degrees
^^divide both sides by cosx
2tanx=1
x=tan^-1(1/2)
tan is positive in the first and third quadrant.
x= 26.565degrees, 206.565degrees
_____________________________________________________________________
The thing that i did not grasp was the identity problems, because i don't get how you can cancel some things out, but not the others. I guess i just have to learn and memorize all of the identities for it to come easier to me, but right now, i have no idea what i'm doing. If someone could show me how to do it I would greatly appreciate it. I just need some examples of how you can cancel the trig functions out and stuff. Because I always want to start with algebra, but you need to start with the identities. It is really tricky, so if anyone wants to make a problem up and work it, i'm going to be your best friendddd. THANKSS.
1. You can not divide trig functions to cancel them
-you can move everything to one side-or-factor out a trig function.
-you can divide by a trig function to create a new one.
_______________________________________________________________________
EX: sinxtanx=3sinx for 0 less than or equal to x less than or equal to 2pi.
sinxtanx-3sinx=0
sinx(tanx-3)<------ factor out a sinx
sinx=0 tanx-3=0
x=sin^-1(0) x=tan^-1(3)
x= 0, pi, 2pi x= 71.565degrees, 251.565degrees
^^you need to use the unit cirlce for this problem because it asks for the inverse of sin of 0, and if you use your unit circle, it is much easier to know it. It conserves time so you don't have to plug it into your caculator.
_______________________________________________________________________
EX: 2sinx=cosx for 0 less than or equal to x less than or equal to 360degrees
^^divide both sides by cosx
2tanx=1
x=tan^-1(1/2)
tan is positive in the first and third quadrant.
x= 26.565degrees, 206.565degrees
_____________________________________________________________________
The thing that i did not grasp was the identity problems, because i don't get how you can cancel some things out, but not the others. I guess i just have to learn and memorize all of the identities for it to come easier to me, but right now, i have no idea what i'm doing. If someone could show me how to do it I would greatly appreciate it. I just need some examples of how you can cancel the trig functions out and stuff. Because I always want to start with algebra, but you need to start with the identities. It is really tricky, so if anyone wants to make a problem up and work it, i'm going to be your best friendddd. THANKSS.
Makeup Reflection #3
These are the graphs that you have to graph wih sin and cos.
y=sin(x) <------ starts on the x-axis
y=cos(x) <------ starts at the max y value
____________________________________________________________________
y=Axin(Bx-h)+C <------- all the same for cos.
*height of wave= A x 2
*A= amplitude
*B= determines perios -----> 2pi/B
**1 max and 1 min ------> how long it takes to repeat.
*h= phase shift (opposite) HORIZONTAL
*c= vertical shift (outside) VERTICAL
___________________________________________________________________
EX: y=2sin(3x+pi)-4
A=2
B=P= 2pi/3
1. 0
2. pi/6
3. pi/3
4. pi/2
5. 2pi/3
**vertical shift = -4
(2/3)/4 = 1/6
h= phase shift = -pi
1. 0-pi= -pi
2. pi/6-pi= -5pi/6
3. pi/3-pi= -2pi/3
4. pi/2-pi= -pi/2
5. 2pi/3-pi= -pi/3
^^add or subtract the phase shift.
^^final x values.
**then you would graph the first one and the second one.
______________________________________________________________
I don't get it when you have to find the equation when they already give you the graph, it confuses me sometimes. If someone could show me a problem like that, i would greatly appreciate it. THANKSS.
y=sin(x) <------ starts on the x-axis
y=cos(x) <------ starts at the max y value
____________________________________________________________________
y=Axin(Bx-h)+C <------- all the same for cos.
*height of wave= A x 2
*A= amplitude
*B= determines perios -----> 2pi/B
**1 max and 1 min ------> how long it takes to repeat.
*h= phase shift (opposite) HORIZONTAL
*c= vertical shift (outside) VERTICAL
___________________________________________________________________
EX: y=2sin(3x+pi)-4
A=2
B=P= 2pi/3
1. 0
2. pi/6
3. pi/3
4. pi/2
5. 2pi/3
**vertical shift = -4
(2/3)/4 = 1/6
h= phase shift = -pi
1. 0-pi= -pi
2. pi/6-pi= -5pi/6
3. pi/3-pi= -2pi/3
4. pi/2-pi= -pi/2
5. 2pi/3-pi= -pi/3
^^add or subtract the phase shift.
^^final x values.
**then you would graph the first one and the second one.
______________________________________________________________
I don't get it when you have to find the equation when they already give you the graph, it confuses me sometimes. If someone could show me a problem like that, i would greatly appreciate it. THANKSS.
Make up Blog Chapter 1-2
This is a review for anyone that needs some refreshing for there exam. I willl be going over a few notes from chapter 1 and 2. In chapter 1 we learned the distance formula. The distance formula is the x2-x1 squared+y2-y1 all square rooted. We did the midpoint formula. To do the midpoint formula you do x1+x2divided by two to get your x and to get your y you do y1+y2 divided by 2.We learned intersections of lines and solving a sysytem of equations. You can solve them two ways:eliminate or substitute. To eliminate you eliminate the variable then solve for the variable, then plug back in. To solve by substituting you solve for a variable, substitute then plug back in. To solve for an x intercept you plug in zero for y and to solve for the y plug in zero for the x. There are three slope formulas. Y-Y1=m(x-x1), y=mx+b, and Ax+By=C. We went over parellel lines and complex numbers. We learned the i chart. This is the i chart=.25=i=square root-1,.5=i2=-1,.75=i3=-i,1=i4=1. Conjugate means the opposite sign of the denomnator. We learned how to complete the square. To do that you move the number to the right, divide by leading coefficient, divide linear term by 2 and square it, add to both sides, factor to left, and solve for x. We learned how to solve quadratics. We can use the quadratic formula and completing the square. We learned how to graph porabalas. The discriminate tells you how many intercepts a graph has. If positve 2 xint, Negative no xint, if 0 1 xint.. Axis of symetry is x=-b/2a. To find the intersection solve for y set equal solve for x, and plug back in. to find the sum you do -2nd over leading coefficent. To find the product you do constant over leading coefficent. A root zero and x intercept are the same thing. Polynomial equations with only addition and subtraction of terms. Leading term is the term with the highest degree. Synthetic division is used to factor. There are special rules to solve anything bigger than a quadratic. Factor by graphing you must have an even number of terms. Quadratic form must have only 3 terms. !st term must=2nd exponent time 2, and last term must be a number. You can also do the rational root therom which takes a long time.
Make Up #5
This week we learned how to solve negative exponents, use logs, and log properties. This thing that I understood the most was how to solve logs.
Example:
Log 2 16=x
2x=16 2x=2^4
X=4
Log 2 16=4
10^x=1020
Log 1020=x
Log 1020=3.009The thing that I don’t under stand from this week was how to solve negative exponents for example (a^-2 + b^2) ^-1/ (1/a^2 + 1/b^2) ^-1. How do you solve this I don’t understand?
Example:
Log 2 16=x
2x=16 2x=2^4
X=4
Log 2 16=4
10^x=1020
Log 1020=x
Log 1020=3.009The thing that I don’t under stand from this week was how to solve negative exponents for example (a^-2 + b^2) ^-1/ (1/a^2 + 1/b^2) ^-1. How do you solve this I don’t understand?
Make Up #4
This week we learned how to solve circles, eclipses, Hyperbolas, and parabolas. The thing that I understood the most this week was how to solve for circles.
Example:
Find the intersection of the circle
X^2+y^2=25 & y=2x-2
1. y=2x-2
2. x^2+(2x-2)^2=25
3. x^2+4x^2-8x+4=25
5x^2-8x+4=25
5x^2-8x-21=0
5x^2-15x+7x-21
(x-3) (5x+7)
X=3 x=-7/5
4. 2(3)-2=4 (3, 4)
2(-7/5)-2=-24/5 (-7/5, -24/5)
The thing that I didn’t understand was how to solve for hyperbolas. I don’t understand how you find the foci.
Example:
Find the intersection of the circle
X^2+y^2=25 & y=2x-2
1. y=2x-2
2. x^2+(2x-2)^2=25
3. x^2+4x^2-8x+4=25
5x^2-8x+4=25
5x^2-8x-21=0
5x^2-15x+7x-21
(x-3) (5x+7)
X=3 x=-7/5
4. 2(3)-2=4 (3, 4)
2(-7/5)-2=-24/5 (-7/5, -24/5)
The thing that I didn’t understand was how to solve for hyperbolas. I don’t understand how you find the foci.
Make Up #3
This week we learned about different kinds triangles, how to solve them, and how to find the area. The thing that I understood the most was using SOHCATOA to solve right triangles.
Example:
If you are given a triangle ABC, side a is 2, side b is 3 and you have to find angle B.
You use tan theta=3/2
Theta = tan^-1(3/2)
Theta = 56.310 degrees
The thing that I didn’t quite understand was using the law of cos. How do you use the formula to solve?
Example:
If you are given a triangle ABC, side a is 2, side b is 3 and you have to find angle B.
You use tan theta=3/2
Theta = tan^-1(3/2)
Theta = 56.310 degrees
The thing that I didn’t quite understand was using the law of cos. How do you use the formula to solve?
Make Up #2
This week we learned how to simplify trigonometry functions, graph trigonometry functions, ideates and equations. The thing I did understood the most was how to graph a trigonometry function. Example:
y=2sin (3x+pie)-4
Amp. =2
Period=2 pie/3 leave pie out p/4=1/6 add to each to each point.
0
Pie/6
Pie/3
Pie/2
2 pie/3
phase shift=pie
0-pie=-pie
Pie/6-pie=-5pie/6
Pie/3-pie=-2 pie/3
Pie/2-pie=-pie/2
2 pie/3-pie=pie/3
Y=2 sin (3x + pie)-4
Then you draw your graph
vertical shift=4The one thing I didn’t understand this week is how to solve a trigonometry function using the identities and equations. I don’t understand how you use the identities and the equations.
y=2sin (3x+pie)-4
Amp. =2
Period=2 pie/3 leave pie out p/4=1/6 add to each to each point.
0
Pie/6
Pie/3
Pie/2
2 pie/3
phase shift=pie
0-pie=-pie
Pie/6-pie=-5pie/6
Pie/3-pie=-2 pie/3
Pie/2-pie=-pie/2
2 pie/3-pie=pie/3
Y=2 sin (3x + pie)-4
Then you draw your graph
vertical shift=4The one thing I didn’t understand this week is how to solve a trigonometry function using the identities and equations. I don’t understand how you use the identities and the equations.
Makeup Reflection #2
I understand mostly what these formulas are. You just have to study and memorize what they are and what problems they go to.
_______________________________________________________________________
Formulas:
1. For a line-
m=tan(alpha) where m=slope
2. For a conic-
tan2(alpha)=B/A-C
3. For a conic if A=C, then alpha= pi/4
__________________________________________________________________
EX: 2x+5y=15 Find the angle of inclination
m= (-2/5) tan(alpha)= (-2/5)
alpha= tan^-1(-2/5)
alpha= 158.199degrees, 338.199degrees
tan is negative in the second and fourth quadrant.
__________________________________________________________________
EX: Find the angle of inclination
x^2 + 3y^2 - 2xy + x = 1
A= 1
B= -2
C= 3
tan2(alpha)= -2/1-3= 1
tan2(alpha) = 1
2alpha= tan^-1 (1)
2alpha= 45degrees, 225degrees
_____________________________________________________________________
The thing that I did not understand was telling what formulas go to each problems. They all look the same to me. All of the x's and numbers, I can not tell the difference between the problems.
_______________________________________________________________________
Formulas:
1. For a line-
m=tan(alpha) where m=slope
2. For a conic-
tan2(alpha)=B/A-C
3. For a conic if A=C, then alpha= pi/4
__________________________________________________________________
EX: 2x+5y=15 Find the angle of inclination
m= (-2/5) tan(alpha)= (-2/5)
alpha= tan^-1(-2/5)
alpha= 158.199degrees, 338.199degrees
tan is negative in the second and fourth quadrant.
__________________________________________________________________
EX: Find the angle of inclination
x^2 + 3y^2 - 2xy + x = 1
A= 1
B= -2
C= 3
tan2(alpha)= -2/1-3= 1
tan2(alpha) = 1
2alpha= tan^-1 (1)
2alpha= 45degrees, 225degrees
_____________________________________________________________________
The thing that I did not understand was telling what formulas go to each problems. They all look the same to me. All of the x's and numbers, I can not tell the difference between the problems.
Makeup Reflection #1
You use these formulas when you are looking for angles. You have to find the inverse of sin, cos, tan, cot, sec, or csc. When you find an inverse, it gives you two angles.
___________________________________________________________________
EX: Find the values of x between 0 and 2pi for which sin= .6
sin is positive in the first and second quadrant
x= sin^-1(.6)
^^you find the inverse of
x= 36.870degrees, 143.130degrees
But the question wants the answer in radians.
So to get radians, you multiply the degrees by pi/180.
You end up with 1229/6000pi, 4771/6000pi.
____________________________________________________________________
EX: 3cosx + 9 = 7 for 0 less than or equal to x less than or equal to 360 degrees.
3cosx= -2
cosx = -2/3
x= cos^-1 (-2/3)
**you do not plug the negative number into your calculator, you would just plug in the number 2/3.
cos is negative in the second and third quadrant.
x= 131.810degrees, 228.190degrees.
____________________________________________________________________
The thing that i did not understand was when you have to find the angle of inclination. I did not get how to use the forulas when you have to find the slope and angle of the line. If someone can help me out with one of these problems, since they might be on the exam, i would appreciate it. Thankss..
___________________________________________________________________
EX: Find the values of x between 0 and 2pi for which sin= .6
sin is positive in the first and second quadrant
x= sin^-1(.6)
^^you find the inverse of
x= 36.870degrees, 143.130degrees
But the question wants the answer in radians.
So to get radians, you multiply the degrees by pi/180.
You end up with 1229/6000pi, 4771/6000pi.
____________________________________________________________________
EX: 3cosx + 9 = 7 for 0 less than or equal to x less than or equal to 360 degrees.
3cosx= -2
cosx = -2/3
x= cos^-1 (-2/3)
**you do not plug the negative number into your calculator, you would just plug in the number 2/3.
cos is negative in the second and third quadrant.
x= 131.810degrees, 228.190degrees.
____________________________________________________________________
The thing that i did not understand was when you have to find the angle of inclination. I did not get how to use the forulas when you have to find the slope and angle of the line. If someone can help me out with one of these problems, since they might be on the exam, i would appreciate it. Thankss..
Make Up #1
This week we learned how to work trigonometry, use the trigonometry chart, the unit circle, and trigonometry inverses. The only thing that I understood this week was how to use the unit circle.
Here is an Example: (-3, 4) find all six trigonometry functions
-draw your graph and draw a line from the origin to the point that is given.
-then use the chart we were given in class and label all six of the trigonometry functions.
-your answer would be:
Sin =4/5
Cos =-3/5
Tan =-4/5
Csc =5/4
Sec =-5/3
Cot =-3/4
The one thing I didn’t understand is how to use the trigonometry chart to solve problems. For example: find the reference angle and the exact answer. How do you determine what to use to solve the problem?
Here is an Example: (-3, 4) find all six trigonometry functions
-draw your graph and draw a line from the origin to the point that is given.
-then use the chart we were given in class and label all six of the trigonometry functions.
-your answer would be:
Sin =4/5
Cos =-3/5
Tan =-4/5
Csc =5/4
Sec =-5/3
Cot =-3/4
The one thing I didn’t understand is how to use the trigonometry chart to solve problems. For example: find the reference angle and the exact answer. How do you determine what to use to solve the problem?
Tuesday, December 8, 2009
Makeup Reflection 4
Ellipse:
(x-h)^2/(length of x/2)^2+(y-k)^2/(length of y/2)^2=1
Center = (h,k)
Major Axis = Larger Denominator
Vertex is on the Major Axis
Focus = smaller # or denominator^2=larger # or denominator^2-focus^2
Steps:
a)Find center
b)Major axis-bigger denominator x or y
c)Vertex + or - square root of big denominator
d)Other integer + or- square root of small denominator
e)Find Focus
f)Length of major axis 2 square root of big denominator
g)Length of the minor axis 2 square root of small denominator
h)Graph
Example:
x^2/25+y/1=1
1. center (0,0)
2.major axis is the x-axis
3.minor axis is the y-axis
4.vertex is (5,0) (-5,0)
5.other denominator is (1,0) (-1,0)
6.the focus is (2 square root of 6, 0) (-2 square root of 6,0)
7.length of major axis 2(square root of 25)=10
8.length of minor axis 2(square root of 1)=2
9.cant do the graph
(x-h)^2/(length of x/2)^2+(y-k)^2/(length of y/2)^2=1
Center = (h,k)
Major Axis = Larger Denominator
Vertex is on the Major Axis
Focus = smaller # or denominator^2=larger # or denominator^2-focus^2
Steps:
a)Find center
b)Major axis-bigger denominator x or y
c)Vertex + or - square root of big denominator
d)Other integer + or- square root of small denominator
e)Find Focus
f)Length of major axis 2 square root of big denominator
g)Length of the minor axis 2 square root of small denominator
h)Graph
Example:
x^2/25+y/1=1
1. center (0,0)
2.major axis is the x-axis
3.minor axis is the y-axis
4.vertex is (5,0) (-5,0)
5.other denominator is (1,0) (-1,0)
6.the focus is (2 square root of 6, 0) (-2 square root of 6,0)
7.length of major axis 2(square root of 25)=10
8.length of minor axis 2(square root of 1)=2
9.cant do the graph
Makeup Reflection 3
Logs and exponentials . So solving logs is easy. Log7^49=x switch the 49 and the x and you will get Log7^x=49 which is obviously x=2 to solve a log that can not be simplified into a whole number such as Log8^23=x You get Log8^x=23 which is xLog8=Log23 so your answer is x=Log23/Log8 . And here is an exponential converted to log form 7^2=49 is Log7^49=2. LOG x^9=2 to solve, switch the 9 and 3 you would get x^2=9 x=3
log 5 of 25 = 2
5^2 = 25
4^3 = 64
log 4 of 64 = 3
log 5 of 25 = 2
5^2 = 25
4^3 = 64
log 4 of 64 = 3
reflection for comments 4
Well this is reflection for comments 4 out of 5. ONLY ONE MORE!!!! Im gonna talk about chapter 7. This chapter will get confusing if you dont memorize a whole lot of formulas and stuff. sin=y/r
tan=y/x
sec=r/x
cos=x/r
cot=x/y
csc=r/y
I know my trig chart like a man though, but im gonna tell you how to convert degrees and radians
really all you have to do for that is multiply the degree by pie/180 and reduce. to change it back to degrees you just multiply it by 180/pie. very simple ay?
tan=y/x
sec=r/x
cos=x/r
cot=x/y
csc=r/y
I know my trig chart like a man though, but im gonna tell you how to convert degrees and radians
really all you have to do for that is multiply the degree by pie/180 and reduce. to change it back to degrees you just multiply it by 180/pie. very simple ay?
reflection for comments 3
Well guys here is 3 out of 5 on my comment reflections haha. Im gonna tell you some stuff about conics and ellipses. I didnt think the stuff we learned was to hard its just trying to remember all of the steps. Here is how to solve an ellipse, here are the steps:
(x-h)^2/(length of x/2)^2+(y-k)^2/(length of y/2)^2=1
Center = (h,k)
Major Axis = Larger Denominator Vertex is on the Major Axis
Focus = smaller # or denominator^2=larger # or denominator^2-focus^2
1)Find center
2)Major axis-bigger denominator x or y
3)Vertex + or - square root of big denominator
4)Other integer + or- square root of small denominator
5)Find Focus
6)Length of major axis 2 square root of big denominator
7)Length of the minor axis 2 square root of small denominator
8)Graph
(x-h)^2/(length of x/2)^2+(y-k)^2/(length of y/2)^2=1
Center = (h,k)
Major Axis = Larger Denominator Vertex is on the Major Axis
Focus = smaller # or denominator^2=larger # or denominator^2-focus^2
1)Find center
2)Major axis-bigger denominator x or y
3)Vertex + or - square root of big denominator
4)Other integer + or- square root of small denominator
5)Find Focus
6)Length of major axis 2 square root of big denominator
7)Length of the minor axis 2 square root of small denominator
8)Graph
reflection for comments 5!
THE LAST REFLECTION FOR COMMENTS!!!!!!! Okay so im going to explain more trigish type thingamajigers... to determine this:
sin (theta) = opposite leg / hypotenuse
cos (theta) = adjacent leg / hypotenuse
tan (theta) = opposite leg / adjacent leg
As well as in plain trigonometry, csc, sec, and cot are found by "flipping" the original values.
Example:
For triangle ABC, a=40, angleA=28 degrees, and angleC is 90 degrees. Solve for b and c.
First you would draw the triangle.
Then take the tangent of 28 degrees, because it deals with the opposite and adjacent sides.
tan 28 degrees = 40/b
b tan 28 degrees = 40
b = 40/tan 28 degrees = 75.229
Then take the sine of 28 degrees, because it deals with the opposite and hypotenuse.
sin 28 degrees = 40/c
c = 40/sin 28 degrees = 85.202
all of this is really easy in general. you really just have to plug everything into the equations for the function and use your calculator. yeah there are times where you need your head but not often. haha. NOW IM DONE WITH MY MAKE UP POSTS!!!
sin (theta) = opposite leg / hypotenuse
cos (theta) = adjacent leg / hypotenuse
tan (theta) = opposite leg / adjacent leg
As well as in plain trigonometry, csc, sec, and cot are found by "flipping" the original values.
Example:
For triangle ABC, a=40, angleA=28 degrees, and angleC is 90 degrees. Solve for b and c.
First you would draw the triangle.
Then take the tangent of 28 degrees, because it deals with the opposite and adjacent sides.
tan 28 degrees = 40/b
b tan 28 degrees = 40
b = 40/tan 28 degrees = 75.229
Then take the sine of 28 degrees, because it deals with the opposite and hypotenuse.
sin 28 degrees = 40/c
c = 40/sin 28 degrees = 85.202
all of this is really easy in general. you really just have to plug everything into the equations for the function and use your calculator. yeah there are times where you need your head but not often. haha. NOW IM DONE WITH MY MAKE UP POSTS!!!
reflection for comments 2
Okay well this is the second reflection of 5 for my missing comments. hahaha. Okay well first off you should know that to find the base of the log all you have to do is look at the number thats under it. Simple right? But if there is not a number, then the base is 10. lets try to put this into exponential form
log base of 2 ^ 4 = xyou would end up with 2^x=4 (that is exponential form)
then you solve
you would set the exponents equal to each other. (x=4)
The easiest thing is when you have to condense and expand the logs. Here is a example of condensing
2log base of 3 ^ x + log base of 1(if it is (+) then you multiply, if it is (-) then you divide)(the number in front of the log becomes an exponent)log base 3^ x^2 log base 12.
now to expand.
log base 2 log base 3/ log base 1log base 2 + log base 3 - log base 1.
Thank you for your time.
log base of 2 ^ 4 = xyou would end up with 2^x=4 (that is exponential form)
then you solve
you would set the exponents equal to each other. (x=4)
The easiest thing is when you have to condense and expand the logs. Here is a example of condensing
2log base of 3 ^ x + log base of 1(if it is (+) then you multiply, if it is (-) then you divide)(the number in front of the log becomes an exponent)log base 3^ x^2 log base 12.
now to expand.
log base 2 log base 3/ log base 1log base 2 + log base 3 - log base 1.
Thank you for your time.
reflection for comment 1
Okay, well this is a reflection to make up some comments. I remember the inequalities and finding the domain and range. I'm going into detail about inequalities.
To solve an inequality, you treat it as a regular equation, but when dividing by a -ve you must switch the sign.
Example:
-2x + 5 lt 7
-2x lt 2
x gt -1 (You switch the sign because you have to divide by -2)
Absolute value inequalities, on the other hand, are a little bit different. You have to come out with two answers. If the initial sign is greater than or = to, or just greater than, it is called an or problem. If the initial sign is less than or = to, or just less than, it is called an and problem.
First, you isolate the absolute value, then you set up two different problems out of the initial problem. One problem is the original. For the second problem, you switch the sign of the value, and switch the sign of the constant on the other side.
Example:
2x + 3 lt 6 (the absolute value is already isolated, so form two equations)
2x + 3 lt 6 and 2x + 3 gt -6
2x lt 3 and 2x gt -9
x lt 3/2 and x gt -9/2
Now before you make that your final answer, you have to plug back in and check. Whichever answers work, you circle or box.
3/2 does not work because if you plug it in, 6 lt 6.
-9/2 does work because -6 lt 6.
So, your answer would simply be -9/2.
To solve an inequality, you treat it as a regular equation, but when dividing by a -ve you must switch the sign.
Example:
-2x + 5 lt 7
-2x lt 2
x gt -1 (You switch the sign because you have to divide by -2)
Absolute value inequalities, on the other hand, are a little bit different. You have to come out with two answers. If the initial sign is greater than or = to, or just greater than, it is called an or problem. If the initial sign is less than or = to, or just less than, it is called an and problem.
First, you isolate the absolute value, then you set up two different problems out of the initial problem. One problem is the original. For the second problem, you switch the sign of the value, and switch the sign of the constant on the other side.
Example:
2x + 3 lt 6 (the absolute value is already isolated, so form two equations)
2x + 3 lt 6 and 2x + 3 gt -6
2x lt 3 and 2x gt -9
x lt 3/2 and x gt -9/2
Now before you make that your final answer, you have to plug back in and check. Whichever answers work, you circle or box.
3/2 does not work because if you plug it in, 6 lt 6.
-9/2 does work because -6 lt 6.
So, your answer would simply be -9/2.
reflection 9
Well this week was pretty simple. You dont really have to worry too much about this. Its really easy to understand it.
For starters, i understood 7.2 the best but changing radians to degrees and vice versa seem like the easiest to explain:
In like all of the problems i have done it makes you change it to radians. Plus if you have to put it in degrees you really need to put the degree symbol or she will mark that beast wrong in a second.
Now, onto converting:
When given radians:
radians TIMES 180/pi.
When given degrees:
degrees TIMES pi/180.
EXAMPLE OF RADIANS:
2pi/8
2pi/8 TIMES 180/pi
2 TIMES 180/8
360/8
45 degrees.
EXAMPLE OF DEGREES:
45 degrees
45 TIMES pi/180
45/180pi
1/4pi
For starters, i understood 7.2 the best but changing radians to degrees and vice versa seem like the easiest to explain:
In like all of the problems i have done it makes you change it to radians. Plus if you have to put it in degrees you really need to put the degree symbol or she will mark that beast wrong in a second.
Now, onto converting:
When given radians:
radians TIMES 180/pi.
When given degrees:
degrees TIMES pi/180.
EXAMPLE OF RADIANS:
2pi/8
2pi/8 TIMES 180/pi
2 TIMES 180/8
360/8
45 degrees.
EXAMPLE OF DEGREES:
45 degrees
45 TIMES pi/180
45/180pi
1/4pi
Once again this week there wasnt really anything i couldnt understand. I hope i kill edee on a test again tho. hahahaha.
reflection 10
This week was all about trigonometry. I dont really like it that much but anyway, we learned a bunch of stuff about trig functions, and I'm gunna help you out with trig functions. Cos, Tan, Csc they're all generally really easy. Study all ways to find trig functions. Period. Then study your trig chart.. A lot. Tonight I'm going to explain how to find all of the trig functions by being told only a point on a graph.
The first thing you need to know if your trig functions and what they are equal to:
sin*= y/r
cos*= x/r
tan*= y/x
csc*= r/y
sec*= r/x
cot*= x/y
*denotes theta
ok now you are given a point...let's say, (3,4)
Now you would usually draw a graph but i can't do that on a computer so here's the shortcut:
Let 3=a and 4=b. On the graph you make a right triangle. So now that you have a and b what are you going to do? That's right, use your formula for to find the hypotenuse of a right triangle: a^2 + b^2 = c^2
So your hypotenuse = 5
now, x=3 y=4 and r=5 because your hypotenuse is r.
Now you plug that into the handy dandy trig functions you memorized. So:
sin=4/5
cos=3/4
tan=4/3
csc=5/4
sec=4/3
cot=3/4
Thats all there really is to it. Its really simple. Get it? Well there isnt too much i dont know from this week. I just really need to study my trig chart. hahaha. well hope you learned something.
The first thing you need to know if your trig functions and what they are equal to:
sin*= y/r
cos*= x/r
tan*= y/x
csc*= r/y
sec*= r/x
cot*= x/y
*denotes theta
ok now you are given a point...let's say, (3,4)
Now you would usually draw a graph but i can't do that on a computer so here's the shortcut:
Let 3=a and 4=b. On the graph you make a right triangle. So now that you have a and b what are you going to do? That's right, use your formula for to find the hypotenuse of a right triangle: a^2 + b^2 = c^2
So your hypotenuse = 5
now, x=3 y=4 and r=5 because your hypotenuse is r.
Now you plug that into the handy dandy trig functions you memorized. So:
sin=4/5
cos=3/4
tan=4/3
csc=5/4
sec=4/3
cot=3/4
Thats all there really is to it. Its really simple. Get it? Well there isnt too much i dont know from this week. I just really need to study my trig chart. hahaha. well hope you learned something.
Love, Chad.
reflection 11
ok, so this week was pretty easy.. Spirit week was pretty cool. I had alot of fun this week. Not really in class because I mean really... who has fun in class. Well let's see...
with the math now.
the sin cos and tan stuff was really easy so lets talk about that.
with the math now.
the sin cos and tan stuff was really easy so lets talk about that.
you use this when you are looking for a missing side of a triangle. Its really simple. You just plug in. you need to put the two things that you know into the formulas below.
cos=adjacent side/hypotenuse
I pretty much understand everything. Im pretty sure im going to do very well on this test. hopefully beating edee:) haha
sin=opposite side/hypotenuse
tan=opposite side/adjacent side
tan=opposite side/adjacent side
so if you have a triangle all you need to do is plug in what it says to find your missing number.
I pretty much understand everything. Im pretty sure im going to do very well on this test. hopefully beating edee:) haha
well hope you learned something at least.
reflection 12
I think that this week of school that we had was pretty much really easy which helps my life out alot. Trig is turning out to be a little easier than i made it out to be in class. I thought it was going to be very hard. I think i can handle this stuff.
Here is a word prolem:
A boy stands 25 feet from the base of a flag pole. He looks up at a 32 degree angle to see the top of the flag pole. How tall is the flagpole?
-ok, you have to now draw this triangle, but i dont know how to do that on here.... draw a right triangle that has a base of 25 feet. Imagine the ground and where the 32 degree angle is, its common sense.
-Now that you've drawn it, use Tan to solve it. Remember SOHCAHTOA. It helps you alot trust me. Tan(angle)=opp/adj. So plug in:
Tan(32)=x/25.
Now solve for x and then you got your answer.
Here is a word prolem:
A boy stands 25 feet from the base of a flag pole. He looks up at a 32 degree angle to see the top of the flag pole. How tall is the flagpole?
-ok, you have to now draw this triangle, but i dont know how to do that on here.... draw a right triangle that has a base of 25 feet. Imagine the ground and where the 32 degree angle is, its common sense.
-Now that you've drawn it, use Tan to solve it. Remember SOHCAHTOA. It helps you alot trust me. Tan(angle)=opp/adj. So plug in:
Tan(32)=x/25.
Now solve for x and then you got your answer.
isoceles triangles still confuse me alot. any help?
reflection 15
Since we didn't have school this week im gonna show you some stuff in Chapter 7 and stuff with the chart and unit circle that we should really learn and not forget.
In 7-1 to convert degrees to radians you have to multipy the degrees by pi/180. To convert radians to degrees you take the radian and multiply by 180/pi.
Ex: 225 degrees x pi/180 = 5/4 pi
3pi/4 x 180/pi=135 degrees
In 7-3 you have to know the unit circle and chart.
The Chart:
Sin=y/r Csc=r/y
Cos=x/r Sec=r/x
tan=y/x Cot=x/y
Ex:
Find all 6 trig functions of (-3,4).
Sin=4/5 Csc=5/4
Cos=-3/5 Sec=-5/3
tan=-4/3 Cot=-3/4
The unit circle:
90 degrees= pi/2 and (0,1)
180 degrees=pi and (-1,0)
270 degrees=3pi/2 and (0,-1)
360 degrees= 2pi and (1,0)
There are some stuff that I need to keep learning but past that its all good.
In 7-1 to convert degrees to radians you have to multipy the degrees by pi/180. To convert radians to degrees you take the radian and multiply by 180/pi.
Ex: 225 degrees x pi/180 = 5/4 pi
3pi/4 x 180/pi=135 degrees
In 7-3 you have to know the unit circle and chart.
The Chart:
Sin=y/r Csc=r/y
Cos=x/r Sec=r/x
tan=y/x Cot=x/y
Ex:
Find all 6 trig functions of (-3,4).
Sin=4/5 Csc=5/4
Cos=-3/5 Sec=-5/3
tan=-4/3 Cot=-3/4
The unit circle:
90 degrees= pi/2 and (0,1)
180 degrees=pi and (-1,0)
270 degrees=3pi/2 and (0,-1)
360 degrees= 2pi and (1,0)
There are some stuff that I need to keep learning but past that its all good.
Monday, December 7, 2009
Makeup Reflection 2
Trig: Angles: from Chapter 7.
Angles are measured in degrees and radians. Angles that are measured in degrees are
measured in degrees minutes' seconds". To find minutes you multiply what is begind
the decimal by 60. To find seconds you multiply what is begind the decimal of
minutes by 60. To reverse the process and find the decimal you divide the minutes by
sixty and add the quotient to the quotient of the seconds over 3600 then add that to
your degrees. Radians= degrees x pi/180 as degrees = radians x pi/180. To find a
coterminal you simply add or subtract 360 degrees or 2pi radians. There are an
infinite number of coterminal angles for all angles.
Sectors:
The formula (S=r theta) is used to find either a radius, angle, or arc length.
(K=1/2r^2) theta is used to find area, a radius, or an angle.
(K= 1/2 rs) is used to find area, a radius or an arc length.
Angles are measured in degrees and radians. Angles that are measured in degrees are
measured in degrees minutes' seconds". To find minutes you multiply what is begind
the decimal by 60. To find seconds you multiply what is begind the decimal of
minutes by 60. To reverse the process and find the decimal you divide the minutes by
sixty and add the quotient to the quotient of the seconds over 3600 then add that to
your degrees. Radians= degrees x pi/180 as degrees = radians x pi/180. To find a
coterminal you simply add or subtract 360 degrees or 2pi radians. There are an
infinite number of coterminal angles for all angles.
Sectors:
The formula (S=r theta) is used to find either a radius, angle, or arc length.
(K=1/2r^2) theta is used to find area, a radius, or an angle.
(K= 1/2 rs) is used to find area, a radius or an arc length.
Reflection 16
I did not get anything that we did in Chapter 8. I know to simplify you have to take the problem and always change some sort of identity before doing Algebra. Identities come before algebra to simplify the problem.
Here are some of the properties
Reciprocal Relationships
csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x
Relationships with Negatives
sin (-x) = -sin xcos (-x) = cos x
csc (-x) = -csc xsec (-x) = sec x
tan (-x) = -tan xcot (-x) = -cot x
Pythagorean Relationships
sin^2 x + cos^2 x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x
These are used to change the problems so they can be done by algebra. I do not know how to do anything else from chapter 8.
Here are some of the properties
Reciprocal Relationships
csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x
Relationships with Negatives
sin (-x) = -sin xcos (-x) = cos x
csc (-x) = -csc xsec (-x) = sec x
tan (-x) = -tan xcot (-x) = -cot x
Pythagorean Relationships
sin^2 x + cos^2 x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x
These are used to change the problems so they can be done by algebra. I do not know how to do anything else from chapter 8.
Reflection 16
So this week the thing I learned the most was identies and the most important thing to learn with identities is to memorize them. If you dont then this stuff is a lot harder to learn. Im actually still memorizing some of them. Here is a way to do it tho.
simplify: tanA x cosA
tan=sin/cos
now its sinA/cosA x cosA and that equals sinAcosA/cosA then your cos cancels, your answer is sinA. Easy right?
here is another one: (1-sinA)(1+sinA)
foil it out with algebra because there is no identities for that 1 - sinA + sinA - sin^2A
=1-sin^2A
Now you have to use a identitiy: sin^2 + cos^2 = 1. solve for 1-sin^2 and you get cos^2 so the answer is really cos^2A.
Identitys can be pretty easy at times but there are some really difficult ones that i cant really understand.
simplify: tanA x cosA
tan=sin/cos
now its sinA/cosA x cosA and that equals sinAcosA/cosA then your cos cancels, your answer is sinA. Easy right?
here is another one: (1-sinA)(1+sinA)
foil it out with algebra because there is no identities for that 1 - sinA + sinA - sin^2A
=1-sin^2A
Now you have to use a identitiy: sin^2 + cos^2 = 1. solve for 1-sin^2 and you get cos^2 so the answer is really cos^2A.
Identitys can be pretty easy at times but there are some really difficult ones that i cant really understand.
I need help with the harder identities and I really need help with the graphing stuff because im totally clueless there.
Makeup Reflection 1
So here i'm refelecting on the law of sines and cosines and the formula for the area of an isosceles triangle.
Law of Sines: sin(opp. angle)/Leg=sin(Opp. angle)/Leg. Set up a proportion.
Example. triangle ABC where A=36 degrees a=3 and B=56 degrees. find b
sin36/3=sin56/x 3sin56/sin36= x Plug into your calculator and done.
Law of Cosines:
(opposite leg)^2=(adjacent leg)^2+(other opposite leg)^2-2(leg)(leg)cos(angle in between)
example:
for a triangle with C=36 degrees a=5 b=6
c^2=5^2+6^2-2(5)(6)cos36
c=Square root of(25+36-2(5)(6)cos36)
c= 3.53
Are of non-right triangle:
1/2(leg)(leg)sin(angle between)
Example: Triangle ABC has sides a=5 b=3 and C=40 degrees
therefore your formula will be 1/2)(5)(3)(sin(40)) type that in your calculator and you will get the answer.
Law of Sines: sin(opp. angle)/Leg=sin(Opp. angle)/Leg. Set up a proportion.
Example. triangle ABC where A=36 degrees a=3 and B=56 degrees. find b
sin36/3=sin56/x 3sin56/sin36= x Plug into your calculator and done.
Law of Cosines:
(opposite leg)^2=(adjacent leg)^2+(other opposite leg)^2-2(leg)(leg)cos(angle in between)
example:
for a triangle with C=36 degrees a=5 b=6
c^2=5^2+6^2-2(5)(6)cos36
c=Square root of(25+36-2(5)(6)cos36)
c= 3.53
Are of non-right triangle:
1/2(leg)(leg)sin(angle between)
Example: Triangle ABC has sides a=5 b=3 and C=40 degrees
therefore your formula will be 1/2)(5)(3)(sin(40)) type that in your calculator and you will get the answer.
Sunday, December 6, 2009
Reflection
ok so we learned identities. Here they are
csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x
sin (-x) = -sin x
cos (-x) = cos x
csc (-x) = -csc x
sec (-x) = sec x
tan (-x) = -tan x
cot (-x) = -cot x
sin^2 x + cos^2 x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x
sin x = cos (90 degrees - x)
tan x = cot (90 degrees - x)
sec x = csc (90 degrees - x)
cos x = sin (90 degrees - x)
cot x = tan (90 degrees - x)
csc x = sec (90 degrees - x)
tan x = sin x/cos x
cot x = cos x/sin x
Always use an identity first. NOT ALGEBRA!!!
csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x
sin (-x) = -sin x
cos (-x) = cos x
csc (-x) = -csc x
sec (-x) = sec x
tan (-x) = -tan x
cot (-x) = -cot x
sin^2 x + cos^2 x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x
sin x = cos (90 degrees - x)
tan x = cot (90 degrees - x)
sec x = csc (90 degrees - x)
cos x = sin (90 degrees - x)
cot x = tan (90 degrees - x)
csc x = sec (90 degrees - x)
tan x = sin x/cos x
cot x = cos x/sin x
Always use an identity first. NOT ALGEBRA!!!
ReFlection 16
This week was pretty good. My weekend went good also. I did alot of hunting but i think i'm starting to get a little sick but i should be ok. Tomorrow we got a math test, God help me!!!!!! Thanks 2 these blogs i'm staying alive. Ive been working on my bridge to so hopefully they boost my grade up by a lot. In math we had to learn some identities. Here are sum of them:cscx=1sinx
Refection #16
This week the thing I grasped most was identies the most important thing to catch with identities is to MEMORIZE them. If you dont you will most definately fail. I'll give an example problem.
simplify: tanA x cosA
because we memorized our identities we know that tan=sin/cos
so now its sinA/cosA x cosA.........which equals sinAcosA/cosA...cos cancels, your answer is sinA. Yay.
another: (1-sinA)(1+sinA)
foil it out....1 - sinA + sinA - sin^2A
=1-sin^2A
once again, our identities give us an equation, which is sin^2 + cos^2 = 1. solve for 1-sin^2 and you get cos^2, so guess what........you're right, the answer to this example is cos^2A.
yet another: sinAcosA
identities let us change cos and we get: sinA cosA/sinA, simplify, and cosA is your answer.
Don't forget to take your time with these problems and remeber, do identities, then algebra, then identities, then algebra, and keep continueing this until it is fully simplified.
____________________________________________________________
I dont get the identities you have to prove. I could use some help with those.
simplify: tanA x cosA
because we memorized our identities we know that tan=sin/cos
so now its sinA/cosA x cosA.........which equals sinAcosA/cosA...cos cancels, your answer is sinA. Yay.
another: (1-sinA)(1+sinA)
foil it out....1 - sinA + sinA - sin^2A
=1-sin^2A
once again, our identities give us an equation, which is sin^2 + cos^2 = 1. solve for 1-sin^2 and you get cos^2, so guess what........you're right, the answer to this example is cos^2A.
yet another: sinAcosA
identities let us change cos and we get: sinA cosA/sinA, simplify, and cosA is your answer.
Don't forget to take your time with these problems and remeber, do identities, then algebra, then identities, then algebra, and keep continueing this until it is fully simplified.
____________________________________________________________
I dont get the identities you have to prove. I could use some help with those.
Reflection #16
This week went by pretty fast, but I have a whole lot of stuff due this week...too much..i probably shouldn't have procrastinated this long...but anyway I had a good week with generally not much homework, which is when I should've been doing the stuff due for this week! Oh well, live and learn. So, to the math...
xsinx tanx = 3sinx for 0 <>
I learned about factoring out trig functions and solving for them.
You can't divide trig functions to cancel them, but you can move everything to one side and factor out a trig function OR you can divide by a trig function to create a new one.
Example:
sinx tanx - 3sinx = 0
sinx (tanx - 3) = 0
sinx = 0
x = sin^-1 (0)
x= 0, pi, 2 pi
tanx - 3 = 0
x = tan^-1 (3)
x = 71.565 degrees, 251.565 degrees
x= 71.565/180 pi, 251.565/180 pi
After determining that the angles need to be in the first and third quadrants, you can then plug the equations into your calculator, and find those inverse angles. Not necessarily will there be just two angles, sometimes there are one, two, three, or four angles to an equation.
What I don't get is #8b and #10 on the Chapter 8 test...help please?
reflection 16
dang, another week gone fast. chapter 8 was the highlight of the week, and the test is tomorrow, bridge due thursday (im working on it), and a video project due friday
identities:
csc X=1/sin X
sec X=1/cos X
cot X=1/tan X
sin (-X)=-sin X
cos (-X)=cos X
csc (-X)=-csc X
sec (-X)=sec X
tan (-X)=-tan X
cot (-X)=-cot X
the chapter is not what i find difficult, but the stuff that needs to be memorized is the challenge
identities:
csc X=1/sin X
sec X=1/cos X
cot X=1/tan X
sin (-X)=-sin X
cos (-X)=cos X
csc (-X)=-csc X
sec (-X)=sec X
tan (-X)=-tan X
cot (-X)=-cot X
the chapter is not what i find difficult, but the stuff that needs to be memorized is the challenge
reflection 16
Finally! The week is over! I hate being sick! and last week was alright, i just love the weekend and how much stuff we have due this week! There's like a million bagillion things due! ahhhhhhhhhhhhh!!!!!!!!!!!!!! Procrastination is only good when there's not alot of stuff due at one time! I found that out the hard way! and omggg!!!!! that bridge!!!! it took a while to do!!!!! but it holds !!!!!!!!!! :)
i hate bein sick!!! i love talking with!!!!!!!!!!!!!!! and i hate/love the norco christmas parade!!!!!!!!!!!! hate: how i gotta march in it while im sick with a cold!!!!!!! love: get to actually c family w/o havin to buy stuff as a gift!!!!!!!! :/
and i hate waking up early on a saturday and sunday!!!!!!!!!!!!!!!
and we had to record a video and we're still not done!!!!!!!!!!!!!!!
aaaaahhhhhhhhh!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
anyways....back to math.....one thing i did understand this week was the whole changing identities and stuff
u kno, like cscø=(1/sinø)
and all that mumbo jumbo
and there was rly nothin else we did this week that i can remember :/
i hate bein sick!!! i love talking with!!!!!!!!!!!!!!! and i hate/love the norco christmas parade!!!!!!!!!!!! hate: how i gotta march in it while im sick with a cold!!!!!!! love: get to actually c family w/o havin to buy stuff as a gift!!!!!!!! :/
and i hate waking up early on a saturday and sunday!!!!!!!!!!!!!!!
and we had to record a video and we're still not done!!!!!!!!!!!!!!!
aaaaahhhhhhhhh!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
anyways....back to math.....one thing i did understand this week was the whole changing identities and stuff
u kno, like cscø=(1/sinø)
and all that mumbo jumbo
and there was rly nothin else we did this week that i can remember :/
Reflection 16
This week went by really fast for some reason, but it's okay cause I liked it:) haha. To me chapter 8 isn't that difficult I just need to remember every step for every different problem. Since I already posted a blog about one thing in chapter 8, I'll explain the next easiest thing. I understand the graphing thing for the most part.
1=A SIN (Bx-h)+c
First you find your amp which is going to be A, or the number before sin the the problem (usually the first number in the problem). Or count the numbers going up and down and divide them by 2. Next find your period, the formula is 2pi/B (B will be the first number in the parenthesis). Once you find that number divide it by 4. Then your going to find 5 numbers, always start with 0 and end with the period number. Add whatever number you got when you divided the period number by 4 to 4 out of the 5 numbers. Then your going to get phase shift numbers. To find the phase shift its the opposite of h. You take the 5 numbers from before this step then you add or subtract(depending on if it's positive or negative) the phase shift from those numbers. Then graph it.
Wow..that was really complicated to explain. It probably makes no sense at all so sorry.
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I pretty much get everything in chapter 8, I just need to review every little thing and memorize all the steps and stuff.
1=A SIN (Bx-h)+c
First you find your amp which is going to be A, or the number before sin the the problem (usually the first number in the problem). Or count the numbers going up and down and divide them by 2. Next find your period, the formula is 2pi/B (B will be the first number in the parenthesis). Once you find that number divide it by 4. Then your going to find 5 numbers, always start with 0 and end with the period number. Add whatever number you got when you divided the period number by 4 to 4 out of the 5 numbers. Then your going to get phase shift numbers. To find the phase shift its the opposite of h. You take the 5 numbers from before this step then you add or subtract(depending on if it's positive or negative) the phase shift from those numbers. Then graph it.
Wow..that was really complicated to explain. It probably makes no sense at all so sorry.
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I pretty much get everything in chapter 8, I just need to review every little thing and memorize all the steps and stuff.
reflection #16
this week went by in a flash... but we spent all week studying chapter 8. one thing i understood was inclination
find the inclination of the line 2x+5y=15
rewrite as y=-2/5x+3
slope=-2/5=tan alpha
alpha=tan^-1(-2/5)=21.8 degrees (reference angle)
180-21.8=158.2 degrees
21.8 degrees, 158.2 degrees
now what i didnt understand was identities... they really confuse me
find the inclination of the line 2x+5y=15
rewrite as y=-2/5x+3
slope=-2/5=tan alpha
alpha=tan^-1(-2/5)=21.8 degrees (reference angle)
180-21.8=158.2 degrees
21.8 degrees, 158.2 degrees
now what i didnt understand was identities... they really confuse me
reflection 16
this week flew by extremely fast. we finished up chapter 8 which i think is pretty confusing. i don't understand 8-4 very well, i guess its because of how you have to memorize the formulas and then do the algebra and then go back to the formulas. it just confuses me. but one thing that i do understand a little is the beginning of chapter 8.
***for example:
find the values of x between 0 and 2pi for which sinx=.6
x=sin-1(.6) sin is positive in the 1st and 2nd quadrants
36.870degrees times pi/180=1223/6000 pi
143.130degrees times pi/180=4771/6000 pi
***another example:
3cosx+9=7 for 0 less than or equal to x less than or equal to 360degrees
3cosx=-2
cosx=-2/3
x= cos-1(-2/3) cos is negative in the 2nd and 3rd quadrants
x=131.810degrees, 228.190degrees
***for example:
find the values of x between 0 and 2pi for which sinx=.6
x=sin-1(.6) sin is positive in the 1st and 2nd quadrants
36.870degrees times pi/180=1223/6000 pi
143.130degrees times pi/180=4771/6000 pi
***another example:
3cosx+9=7 for 0 less than or equal to x less than or equal to 360degrees
3cosx=-2
cosx=-2/3
x= cos-1(-2/3) cos is negative in the 2nd and 3rd quadrants
x=131.810degrees, 228.190degrees
reflection 16
Hey everybody!!! THE SAINTS ARE 12-0!!! THE COLTS ARE 12-0!!!! woop woop!!! LSU is goin to the capital one bowl!!!! I just got home from the christmas parade in norco, which was very nerve wrecking because I wasn't watching the saints game or the colts game! Not as much the colts though, man i thought the saints were gonna lose, but I'm super pumped they won!!!!
So, now on to the math! Almost everything we did this week had to do with identities! Which is super stupid because i don't like remembering things! lol!
So these are just some of the many identities we have learned:
csc A = 1/sin A
sec A = 1/cos A
cot A = 1/tan A
sin (-A) = -sin A
cos (-A) = cos A
csc (-A) = -csc A
sec (-A) = sec A
tan (-A) = -tan A
cot (-A) = -cot A
sec A = 1/cos A
cot A = 1/tan A
sin (-A) = -sin A
cos (-A) = cos A
csc (-A) = -csc A
sec (-A) = sec A
tan (-A) = -tan A
cot (-A) = -cot A
________________________________________________________
Now, for what I don't know....
Does anyone wanna help me on knowing when to use algebra or identities???
It keeps confusing me!
Reflection
Well i'm about to do another blog, and i am starting to run out of things to say on these things so i'll make something up. We been working on Chapter 8 and tomorrow is the chapter 8 test. I know most of the work but i'm still not positive about everything. We learned about identities, amplitudes, periods, and graphing those, and we are still resorting back to the trig chart.
EXAMPLE:
5 cos theta = -1 for 0 degrees < or equal to theta < 360 degrees
divide 5 divide 5
cos theta = -1/5
checks in the 2nd and 3rd quadrin
= 101.537 degrees, 191.573 degrees
That's basically what i understood the most for chapter 8.
EXAMPLE:
5 cos theta = -1 for 0 degrees < or equal to theta < 360 degrees
divide 5 divide 5
cos theta = -1/5
checks in the 2nd and 3rd quadrin
= 101.537 degrees, 191.573 degrees
That's basically what i understood the most for chapter 8.
Reflection #16
This week, I understood most of it. One of those things is...
Identities:
csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x
sin (-x) = -sin x
cos (-x) = cos x
csc (-x) = -csc x
sec (-x) = sec x
tan (-x) = -tan x
cot (-x) = -cot x
sin^2 x + cos^2 x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x
sin x = cos (90 degrees - x)
tan x = cot (90 degrees - x)
sec x = csc (90 degrees - x)
cos x = sin (90 degrees - x)
cot x = tan (90 degrees - x)
csc x = sec (90 degrees - x)
tan x = sin x/cos x
cot x = cos x/sin x
**x can also be said as alpha, theta, A, B, etc.
Examples:
Simplify: sec x - sin x tan x
1/cos x - sin x/1 (sin x/cos x)
1/cos x - sin^2 x/cos x
1-sin^2 x/cos x
cos^2 x/cos x
= cos x
tan x times cos x
sin x/cos x times cos x/1
= sin x
(1 - sin x)(1 + sin x)
1 - sin^2 x
=cos^2 x
csc x - sin x
1/sin x - sin x/1 (sin x/sin x)
1 - sin^2 x/sin x
cos^2 x/sin x
= cot x cos x
Now, what I don't understand...
Chapter 8 test (page 329) # 8d and #10b
I tried, but I got lost. Could anyone help explain these problems?
Identities:
csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x
sin (-x) = -sin x
cos (-x) = cos x
csc (-x) = -csc x
sec (-x) = sec x
tan (-x) = -tan x
cot (-x) = -cot x
sin^2 x + cos^2 x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x
sin x = cos (90 degrees - x)
tan x = cot (90 degrees - x)
sec x = csc (90 degrees - x)
cos x = sin (90 degrees - x)
cot x = tan (90 degrees - x)
csc x = sec (90 degrees - x)
tan x = sin x/cos x
cot x = cos x/sin x
**x can also be said as alpha, theta, A, B, etc.
Examples:
Simplify: sec x - sin x tan x
1/cos x - sin x/1 (sin x/cos x)
1/cos x - sin^2 x/cos x
1-sin^2 x/cos x
cos^2 x/cos x
= cos x
tan x times cos x
sin x/cos x times cos x/1
= sin x
(1 - sin x)(1 + sin x)
1 - sin^2 x
=cos^2 x
csc x - sin x
1/sin x - sin x/1 (sin x/sin x)
1 - sin^2 x/sin x
cos^2 x/sin x
= cot x cos x
Now, what I don't understand...
Chapter 8 test (page 329) # 8d and #10b
I tried, but I got lost. Could anyone help explain these problems?
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