Reflection #19

This week, I'm going to review the areas of different triangles:

Area of a right triangle: A=(1/2)(base)(height)
Area of a non-right triangle: A=(1/2)(leg)(other leg)sin(angle in between the two legs)

Examples:
visualize a right triangle: DEF (left to right)
D=90 degrees, d=8, e=6...Find the area.
So, you know your base is e, which is 6.
Now, to find your height, or f, use: a^2 + b^2 = c^2
c is always to hypotenuse, which is d (8).
6^2 + b^2 = 8^2
36 + b^2 = 64
b^2 = 28
b = 5.292
Your height is 5.292
Now, plug into the formula.
A = (1/2)(6)(5.292)
A = 15.875

Now, visualize a non-right triangle: HIJ (left to right)
H = 65 degrees, j = 2, i = 6
Plug into the formula.
A = (1/2)(2)(6)sin(65)
A = 5.438

Holiday Reflection

So heres a refelection 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.

REFLECTION #19

Time for another blog! yay. Hope everyone is enjoying their holidays because I'm not (: Anyway, I think I'll just review the first things that pop into my head. This should be quite interesting...

Well, let's see I think way back in like Chapter 3 or something we learned inequalities so I'll start off with something easy like that. hah. To start off, there are two types of inequalities, regular inequalities and absolute value inequaties. In absolute value inequalities there are two different types determined by the signs if it is not an equal sign. They can either be "and" or "or" inequalities. "And" inequalities always have a less than symbol (<) and "or" inequalities always have a greater than symbol (>). Also, (something important to remember) in absolute value inequalities you always get two answers. I'm sure everyone remembers how to work these problems but here's a few examples:

Ex. 1.) 4x + 1 > 13
*you would just solve this like a regular equation even though it does not have an equal sign. *and for this problem you would only get one answer because it does not have an absolute value symbol.
*so you subtract 1 from 13 and you get 4x > 12
*divide by 4 and you get that x > 3

Ex. 2.) 2y - 4 = 12 (*assume there is an absolute value thing around 2y-4)
*the first thing you want to do with this problem is set up two equations. since in absolute value you know you will get two answers
*your first equation will be the original equation but without the absolute value symbol like this:
2y - 4 = 12
*and your second equation will be the same equation except you change the 12 to -12 like this:
2y - 4 = -12
*then you solve both equations for y and you get that y = -4 and 8

Ex. 3.) 3x - 4 + 5 < 27 (*assume there is an absolute value thing around only 3x - 4)
*the first thing you have to do is isolate the absolute value and to do that you have to get rid of the 5. so you subtract the 5 over from the 27 and your new equation is:
3x - 4 < 22
*the next thing you notice is that this equation has a < sign and that means it is an "and" equation so you have to set it up a certain way like this:
-22 < 3x - 4 < 22
*then you solve the equation first by adding 4 over to all sides then you get this:
-18 < 3x < 26
*then you divide by 3 on all sides
your final answer is: -6 < x < 26/3

Next I think I will go over slope. First of all, there are 3 different equations of slope. There's slope intercept form, point slope form, and standard form. Slope intercept is y = mx + b. Point slope form is y - y1 = m (x - x1). And standard form is Ax + By = C
Here are a few examples of problems:

Ex. 4.) Find the slope of the two points (4,1) (3,0)
*to solve this you use the formula for slope which is: m = y2 - y1/x2 - x1
*so you get...0 - 1/3 - 4 and that gives you -1/-1 which equals 1

Ex. 5.) Find the equation of the line 3x + 4y = 12 that is perpendicular to the point (3,2).
(*by the way, I'm not sure if I worded that problem right but hopefully you should know what I'm talking about right?)
*so the first thing you have to do is find the slope of the equation you are given. 3x+4y=12
*first you subtract 3x over and you get 4y = -3x + 12
*then you divide by 4. and you get that y = -3/4x + 3
*So your perpendicular slope of that equation is 4/3. (because you take the negative reciprocal of the orginal slope of the equation)
*Then to put it in an equation including the point you are given, you use the point slope formula.
*So your final answer is y - 2 = 4/3 (x - 3)

Umm..the last thing I can think of at the moment is reference angles and exact answers. *Remember that reference angles can only be between 0 and 90 degrees. Here are some examples:

Ex. 6.) Find the reference angle of sin 236. (*assuming that there's a degree sign after 236)
*the first thing you have to do is figure out what quadrant 236 is in. it's in the 3rd quadrant.
*sin relates to the y axis and in the third quadrant, the y axis is negative so for the reference angle, sin has to be negative
*and to find the reference angle of 236 you have to subtract 180. and you get 56 degrees.
*so your reference angle is equal to -sin 56. and since this is not on the trig chart, to find the exact answer you would have to type it into your calculator.

Okay well I think I reviewed enough for this week. ha

reflection 19

This week over the holidays im going to review chapter 9, which is right triangles. There are 3 things that you have to remember: hypotenuse opposite, A=1/2bh, and SOHCAHTOA, which is:

sin=opposite leg/hypotenuse

cos=adjacent leg/hypotenuse

tan=opposite leg/adjacent leg

Here are some examples on using SOHCAHTOA....

solve for b & c.

1) You have an ABC triangle with angle A 28 degrees, a=40, and angle C is a right triangle.
tan28=40/b
btan28=40
b=40/tan28

which equals about 75.229

sin28=40/c
c=40/sin28

which equals about 85.202

Reflection 19

Today over the holidays i'm going to review the trig chart and the unit circle, i bet remembering a few of those could of helped me get a few more points on my exam.
Wow its amazing how much or little i forgot of that stuff.
Now the steps in solving an inverse is used to solve for an angle, you cant divdie by a trig function. An inverse will find 2 angles.
The steps determine the quad of your two angles, use the trig chart to find a reference angle, if not on chart or unit circle use the calcualter.
S=rtheta, remember that. SOH CAH TOA is a very helpful tool in solving right triangles. Law of sines is sina/a=sinb/b=sinc/c. Keep enjoying your holidays!!!!!!!!!!!!!!!!!!!!!

Reflection 18

Yea school is over and we get our long two week break. I am looking forward for this break cause i need to have a break. I plan to do a lot of hunting and fishing, and chilling with my friends. Ive been working on a bonfire as well. Christmas night went well to, i enjoyed myself alot and my break has been doing a lot of good for me. But to keep my brain fresh in mind i will review a few math notes that we took over the first semester. To find the intersection you must put them equal to. Example:y=2x+5 y=8-x^2 2x+5=8-x^2 x^2=2x-3=0 (x+3)(x-1)=0 so x=1,-3.
To sketch polynomial function you must first factor completely, set up a # line lable zeros, plug in on either side of your root, posisitve intergers above x-axis, negative below x-axis, check in calculater, max in min calc only you find the vertex which is -b/2a. I also though the exam was kinda haed but then it was easy i just 4got how to do a bunch a simple problems that i'm kicking myself in the ass for.

Reflection

so i hope everyone had a great christmas and i guess we were supposed to do another blog but i dont know. Here is some review stuff from chapter 8. We learned how to find the angle of inclination, which i found was really easy compared to some stuff we learn. We also learned about amplitudes, periods, vertical shifts, etc. There are some formulas we had to learn to be able to work these problems:

1.) For a line
m=tan alpha where m=slope and alpha=angle of inclination
2.) For a conic
tan 2 alpha=B/A-C
3.) For a conic if A=C then
a=pi/4

EXAMPLE:

Find the angle of inclination.
2x+5y=15
m=-2/5 tan alpha=-2/5 Checks are in the II and IV area and 21.801 degrees in I
alpha=tan^-1(-2/5)
180-21.801 alpha ~ 158.199 degrees, 338.199 degrees
158.199+180

reflection 18

alright, i forgot to do this sunday..so this is the blog for december 20. soo...that exam, pretty hard...but im so glad that we are off for the christmas holidays! i guess i will do my blog on 7-3, this stuff i actually understand and i think it is pretty easy. all you have to remember is the little chart. which is:

sin= y/r
cos=x/r
tan=y/x

csc=r/y
sec=r/x
cot=x/y


and to find r it is the square root of x^2 + y^2.

this chart is really easy, all you have to remember is the first part and then the second part is just the opposite.
and there is also the unit circle that goes along with this, but i can't really draw that on here..

heres an example:

find all six trig funtions (1,-1)

sin=-square root of 2/2
cos=square root of 2/2
tan=-1
csc=square root of 2
sec=square root of 2
cot=-1

REFLECTION #18

Okay so since I don't have my notes with me I'll just go over some simple things that we learned early this year. We should all know how to do these really good by now. ha. I'll explain the different types of factoring.

First there's factoring by using simple algebra. Here's an example of that:

Ex. 1.) 2x^2 + 4x = 6
*the first thing you notice that you can do is factor out a 2x like this:
2x(x + 2) = 6 >>(and what's in parenthesis is what you get after you divide by 2x)
*then you take both things and set them equal to 6 and then solve:
2x = 6 >> = 3
x + 2 = 6 >> = 4
*So your final answers in point form are (3,0) (4,0)

Next there's factoring....(I don't know exactly what it's called so I'll just say it's regular factoring)
Here's an example of those:

Ex. 2.) x^2 + 7x + 12
*to factor this equation you can say, "what are the factors of 12 that add to give me 7?"
*factors of 12 that add to give you 7 are 4 and 3
*so since both signs are addition you write it as >> (x + 4)(x + 3)
*then you solve for x by setting them both equal to zero. x + 4 = 0 x + 3 = 0
*And your final answers in point form are (-4,0) (-3,0)

Ex. 3.) 2x^2 + 5x + 3 = 0
*to factor this you have to first multiply the outer numbers (2 and 3) and you get 6. Then you can ask yourself "what are the factors of 6 that add to give me 5?"
*that would be the numbers 3 and 2
*so you rewrite your problem like this: 2x^2 + 3x + 2x + 3 = 0
*now since you have 4 terms, you can group them like this: (2x^2 + 3x) + (2x + 3) = 0
*now solve. you can take an x out of the first parenthesis. so you get x(2x + 3) + 1(2x + 3) = 0
*then you get (2x + 3)(x + 1)
*then set them both equal to zero and solve for x.
2x + 3 = 0 >> 2x = -3 >> x = -3/2
x + 1 = 0 >> x = -1
*Your final answers are (-1,0)(-3/2,0)

The other two ways to factor are the quadratic formula and completing the square. You use the quadratic formula when regular factoring doesn't work and when the linear term is odd. You use completing the square when the linear term is even.

Here's a few examples:

Ex. 4.) Complete the square. x^2 + 8x + 9 = 0
*the first thing you want to do is move the 9 over to the other side. so you subtract 9.
x^2 + 8x = -9
*then you leave a space between the 8x and the equal sign like this:
x^2 + 8x = -9
*then you take the linear term, which is 8x, and divide it by 2 and square it. so you get 16.
*then you add 16 to both sides of the equation like this:
x^2 + 8x + 16 = -9 + 16 >> then simplify it and get this:
(x + 4)^2 = 7
*then take the square root of both sides and you get this:
x + 4 = +/- square root of 7
*then subtract 4 over and get x = -4 +/- square root of 7

Now for the quadratic formula!! woot woot.

Ex. 5.) 2x^2 + 6x + 5 = 0

*the formula you use is x = -b +/- square root of b^2-4ac/2a
*so when you plug the numbers into the formula you get x = -6 +/- square root of 36-4(2)(5)/4
*then simplifying that you get: x = -6 +/- square root of -4/4
*and simplifying it again you get: x = -3 +/- i over 2

And there yah go!!!! (:
MERRY CHRISTMAS EVERYBODY (:

reflection

Im gonna show you something in chapter 2. Solving anything bigger than a quadratic usuing quadratic form. To use quadratic form you have to have 3 terms only. The first term must equal the 2nd exponentx2, and the last term must be a constant. The first thing you do is make g=x^exponent/2 so that you would get g^2+g+#. The second thing is to do the quadratic formula, factor, or complete the square. The the last thing is to plug back in for g. (Whenever you do step three you are basically just plugging back into step one. g=x^2)

An example:
x^4-4x^2-12=0

1. g=x^4/2
g=x^2
g^2-4g-12

2. (g^2-6g)+(2g-12)
g(g-6)+2(g-6)
(g+2)(g-6)
g=-2 g=6

have a good christmas guys:)

Reflection #18

Okay, how about a review on the rational root therom:

ex. x^4+2x^3-7x^2-8x+12
factors of p (constant) = +/-12, +/-1, +/-6, +/-2, +/-3, +/-4
factors of q (leading coeff.) = +/-1
find p/q=+/-12, +/-1, +/-6, +/-2, +/-3, +/-4
plug all possibilities into table on calculator and find all where y is 0.
you get 1, 2, -2, and -3.
do synthetic division with any two.
1 1 2 -7 -8 12
1 3 -4 -12
1 3 -4 -12 0
(x^3+3x^2-4x-12)(x-1)
2 1 3 -4 -12
2 10 12
1 5 6 0
(x^2+5x+6)(x-1)(x-2)
solve by factoring x^2+5x+6=
(x+2)(x+3)
x=-2, -3, 1, 2
put in point form:
(-2, 0)(-3, 0)(1, 0)(2, 0)

Simply stuff from chapter 2(?).

reflection

HAPPY HOLIDAYS! Hope everyone has a Merry Christmas and Happy New Year!!



A little review of Chapter 7.


section one.

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 the reciprical of how you convert to radians 180/pi.

Ex: 225 degrees x pi/180 = 5/4 pi
3pi/4 x 180/pi=135 degrees

section three.


Know the unit circle and chart. STUDYYYY!



The Chart: find all 6 trig functions.
Sin=y/r Csc=r/y
Cos=x/r Sec=r/x
tan=y/x Cot=x/y

EXAMPLE:

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



andd 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)

that can be used when finding coterminal angles.

Reflection

Ok. Now we have trig involving lines and conics. The basic formula for a line in this section is m=tan(alpha)
m=slope
(alpha)= angle of inclination.

And don't forget that when you are finding an inverse that you are finding two angles. And don't forget that you must also check the quadrants. The positive quadrants for tangent are one and three. Making two and four the negative quadrants for tangent.

I'll use an example she gave us in class: 2x+5y=15 find the angle of inclination (alpha). Find your slope. Slope=-A/B so that is -2/5 referring to the standard formula Ax+By=C. now (-2/5)=tan(alpha) since we dont divide by trig functions you take an inverse of tangent to find your angle. With your calculator the inverse tan of -2/5 is -21.801. Ignore the negative so that you have 21.801. We need this answer in the quadrants where tan is negative. Because our slope was negative. To move 21.801 to quadrant two you make it negative then add 180. To move it to the fourth quadrant you make it negative then add 360. Those are your two angles of inclination and equal alpha.

The basic formula for a conic is tan(2x(alpha))=B/(A-C)

To make things easier you should remember that if A=C then (alpha)=Pi/4

The basic formula for a conic is Ax^2 + Bxy +Cy^2 +Dx + Ey + F=0
you should know that

Reflection 18

Okay, so I'm finally glad we're all done exams, and I'm also glad we're on Christmas break! This week we basically did nothing but take exams, and on Monday we had a review day, and we turned in our study guides. Finding symmetry is pretty easy, and I'm going to explain that.

To find symmetry on the y-axis:
you have to plug in (-x) to every x there is

Example:
y = x^2 - 7
y = (-x)^2 - 7
y = x^2 - 7
Therefore, yes, there is symmetry.

To find symmetry on the x-axis:
you have to plug in (-y) to every y there is

Example:
y = x - 2
(-y) = x - 2
y = -x + 2
Therefore, no, there is no symmetry.

To find symmetry on the origin:
you have to plug in (-y) and (-x)

Example:
y = x^2 + 4
(-y) = (-x)^2 + 4
y = -x^2 + 4
Therefore, no, there is no symmetry.

To find symmetry on y=x:
find the inverse

Example:
y = x^2 + 4
x = y^2 + 4
y^2 = x - 4
y = sqrt(x-4)
Therefore, no, there is no symmetry.

Reflection 18?

YESSSSSSSSSSS! We are finally off for the holidays and it's about time cause i'm so tired of school. Two weeks of no math :)..anyways the exam was super hard and I probably failed. But back to this blog since we didn't really learn anything something I remembered how to do was logs. Logs are pretty easy you just have to remember all the rules. Remember that if theres a minus sign in the problem your gonna divide and if theres a plus sign your going to add.

To condense a log-Example:

log 4 - log pi + 3 log r - log b

*You can pull the log out of this, and remember -/divide, +/multiply.

Answer: log(4r^3/pi b)

**Remember when theres a number infront of a log you raise it to the number or variable behind the log. ex: 3 log r...log r^3.

------------------------------------------

Since were off and I don't have anything with me, I can't think of anything specific that I don't understand.

Reflection

WOW! What a week. One word explains it, EXAMS! This week just review over stuff from chapters 1-8, which is a lot of information to review. I'll explain again from when we learned about domain, range, functions, finding any symmetry, and inverses. What I found was the easiest to learn was finding the symmetry.

EXAMPLE: y=x^3=4x

a.) x-axis
(-y)=x^3+4x
y=-x^3-4x Reflect; No

b.) y-axis
y=(-x^3)+4(-x)
y=-x^3-4x No

c.) y=x
x=y^3+4y
y^3+4y=x
y(y^2+4)=x No

d.) origin
(-y)=(-x)^3+4(-x)
-y=-x^3-4x
y=x^3+4x Yes

I also understood how to do domain and range very well. The types of the domain and range problems i understood the most were the fractions and polynomials. (oo => stands for infiniti

EXAMPLE: y=x^3+4x^2+12
For any type of polynomial the domain would be (-oo,oo) and for the range, odd:
(-oo,oo); quadratics: [vertex,oo] or [-oo,vertex].
The answer to this problem would be Domain: (-oo,oo) Range (-oo,oo)

EXAMPLE: 5x+4/x^2-4
First, set the bottom of the fraction equal to zero.
x^2-4=0
+4+4
x^2=4
x=+ or - 2
Your answer then comes to,
Domain: (-oo,-2)u(-2,2)u(2,oo)

So thats blog number one for the holidays. I think what i had the most trouble with on the exam was from chap. 9.

Reflection #18

WOW! that exam was really, really easy..if you studied. It think i did really good. I did all of my study guides and i understood everythign that was on them, so...and one thing that i did understand on the exam was how to find the symmetry. And by the way, the exam was on chapters 1-8. AHHHHH. PS: just because i said i think i did good on my exam, i probably failed, lol.

___________________________________________________________

EXAMPLES:


y=x^3=4x


a.) x-axis
(-y)=x^3+4x
y=-x^3-4x Reflect; No

b.) y-axis
y=(-x^3)+4(-x)
y=-x^3-4x No

c.) y=x
x=y^3+4y
y^3+4y=x
y(y^2+4)=x No

d.) origin
(-y)=(-x)^3+4(-x)
-y=-x^3-4x
y=x^3+4x Yes


^^the only thing this problem was symmetric about was the origin

_________________________________________________________________


I still do not understand how to do the domain and range. I know how to do that number line thing, but when i do not understand how to do a problem, i just put (-infinity, infinity). I know that for the fraction problems, youi just use the top part, but if someone can give me an example of the domain and range problems, i would greatly appreciate it. I really need help with these problems. And we will probably need these lessons as we go further anyway, so i might as well learn it now. THANKSSS :)

Relfection #17

This week we've been working on study guides. I'll reflect about something we learned a couple weeks ago. This might help you on your study guides. This is how you find the area of a triangle when given two lengths and angle.

For Area say you have a triangle with:

A side length of 4
A side length of 5
And angle of 30 degrees

Then plug it all in: A=1/2 (4) (5) Sin 30 degrees

A= 10 Sin 30 degrees which aproximately =5.

For another triangle:

A side length of 3
A side length of 8
And an angle of 60 degrees

Then plug it all in: A=1/2 (3) (8) Sin 60 degrees

A=12 Sin 60 degrees which approximately = 10.392

Reflection #17

This week we worked on study guides. From chapters past, one thing I'm going to reflect on is logs: change of base:

**Only use when there is no other way to change the base of a log or solve for x in an exponential
First, write as an exponential if it is not already
Second, take the log of both sides (base 10)
Third, move exponent (x) to front
Forth, solve for variable by division
Fifth, write as log fraction (simplify further if whole number only, plug in calc.)

Example:
log base 3 of 7
3^x=7
log of 3^x=log of 7
x log of 3=log of 7
x=log of 7/log of 3

5^x=10
log of 5^10=log of 10
**Note: log of 10 = 1
x log of 5=1
x=1/log of 5

Now, I don't understand from the study guides...
Chapter 8 - #4, #9, #10, #11
Chapter 7 - #16, #17, #21
Any help with these problems??

Reflection 17

Last week we really didnt do too much. It was a pretty easy, but busy week. We worked on our study guides and did our bridge project. Unfortunatly some people had to do it twice, and we still got the wrong glue, thanks to who ever gave us that. Anyways, lets review about right angles of triangles to help with our mid term exam that is coming up on wednesday. Right Triangle, here we go!!!! The hypotenuse is always opposite a right angle. The area of a right triangle is 1/2 the base x height. To deal with traingles remember the saying soh cah toa. This saying means sin is equal to the opposite leg divided by the hypotenuse, cos is equal to the adjacent leg over the hypotenuse, and tan is equal to the opposite leg divided by the adjacent leg. You always want your calculater in degrees when trying to solve these problems. When you want the degree of a triangles angle you have to hit the second button first so that u get the inverse, also knowing the trig chart will help on some of those problems, so just a hint, KNOW THE TRIG CHART< IT WILL HELP!!!!!!!!!!!!!! Finding the area of a non right traingle is pretty easy all you have to do is A=1/2(leg)(leg)sin(angle). We also learned the law of sin and we learned the law of cosine, i was absent those days so anyone that wants to show me an example prblem and help me i would appreciate that, especially the law of cosine, i dont even think i have the notes.

Reflection 17

I did not know what to reflect about from this week since we did study guides and our bridge contest. So I decided to reflect about something we learned a couple weeks ago. This might help you on your study guides. This is how you find the area of a triangle when given two lengths and angle.

For Area say you have a triangle with:

A side length of 4
A side length of 5
And angle of 30 degrees

Then plug it all in: A=1/2 (4) (5) Sin 30 degrees

A= 10 Sin 30 degrees which aproximately =5.

For another triangle:

A side length of 3
A side length of 8
And an angle of 60 degrees

Then plug it all in: A=1/2 (3) (8) Sin 60 degrees

A=12 Sin 60 degrees which aproximately = 10.392

This formula can only be done when you have two given lengths and an angle.

Reflection

This week was mostly study guides in advanced math so nothing really new. I do however need some help with simplifying trig functions so if anyone would help me with an example that would be nice. So far i think that tan is equal to sin over cos and i know that sin squared plus cos squared always equals one. And that tan is equal to one over sin and cot is equal to one over cos. Could somebody please post a few more examples before the exam.

Reflection 17!

So this is going to be a little harder than usual because we didn't really learn anything new this week. Our bridge held up to 75 pounds, how cool? :) Congrats to the winning bridge, 20 points sounds like a great Christmas present! Keenen, we'll alll miss you. Don't forget to keep in touch, and any chance that you get to come visit PLEASE DO! We loooove you! Good luck wherever you go, :/

I'll just go through my study guides and try to explain a few things from a while back?
First, i'll start with synthetic division.
Synthetic division is strictly used to find zeroes or roots of polynomials.

Equation: (x^5+x^4+2x^3+5x^2-3x+6) / (x+3)
First you'll set x+3=0
x=-3

this gives you what to synthetically divide by(not sure if that's the correct way to say that! lol) :

-3|1 1 2 5 -3 6




1st: Bring your first number down.

-3|1 1 2 5 -3 6
------------------
1

2nd: multiply the number at the bottom by the number being divided and add.

-3|1 1 2 5 -3 6
| -3 6 -24 57 -162
---------------------------
1 -2 8 -19 54 -156

Remainder for this equation would be:
R= -156.

All in all,
these are the steps in words...

• Bring down the 1.
• Multiply it by the -3 to get -3.
• Add 1+-3 to get -2.
• Multiply -2 times -3 to get 6.
• Add 2 plus 6 to get 8.
• Multiply 8 times -3 to get -24.
• Add 5 plus -24 to get -19.
• Multiply -19 times -3 to get 57.
• Add -3 plus 57 to get 54.
• Multiply 54 times -3 to get -162.
• Add 6 plus -162 to get -156.

Second, the six trig functions are:

sin=y/r
cos=x/r
tan=y/x
cot=x/y
csc=r/y
sec=r/x

ALSO:
degrees to radians= D * pi/180
radians to degrees= R * 180/pi(in this case, pi's cancel leaving you with a number in DEGREES)

and finally, exponential equations?

in an equation:
(b^2/a)^-2 TIMES (a^2/b)^-3

you must first distribute your exponents to all parts of your fraction, in the case of exponents, you must multiply the exponential values together, i.e.,

(b^-4/a^-2) TIMES (a^-6/b^-3)

in order to remove all negative exponents from the equation, put a 1 over each and use the sandwich method!

(1/b^4)/(1/a^2) TIMES (1/a^6)/(1/b^3)

multiply the top by the bottom(BREAD) and the two inner fractions by each other(MEAT, or peanut butter- whichever you prefer. lol)

a^2/b^4 TIMES b^3/a^6

cancel what you can:

a^2 and a^6= a^4
b^4 and b^3= b


FINAL EQUATION:

1/ba^4

---------------------------------------------------------------
WORD PROBLEMS STILL GET ME, PLEASE PLEASE PLEASE HELPPP!

Reflection 17

So, our bridge won! 20 bonus points i hope! So this week what we did was our study guides and bridges, which we had to stay up till 430 in the morning! So these are the six trig functions and the unit circle:

Six trig functions:
sin: y/r
cos: x/r
tan: y/x
cot: x/y
csc: r/y
sec: r/x

Unit Circle:
90(pi/2): (0,1)
180(pi): (-1,0)
270(3pi/2): (0,-1)
360(2pi): (1,0)

And here are a couple 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

and thats all the time i have folks! peace!

reflection 9

The thing we learned this week was identities. In identities you pretty much just keep canceling things out. The thing you have to remember in identities is first you find an identity then you do some type of math to make the problem smaller. In this you have to memorize all of the identites it's the only way you will be able to work the problems.

These are the identites you will use a lot:

Pythagorean Relationships:

sin^2 x + cos^2 x = 1

1 + tan^2 x = sec^2 x

1 + cot^2 x = csc^2 x

Reciprocal Relationships:

csc x = 1/sin x

sec x = 1/cos x

cot x = 1/tan x

Ill give an example.

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.

Identities are chap and you have to be all careful and stuff if you wanna do it right. just remember to do identities before your math stuff and you straight.

I forgot like graphing. pretty much everything about it. aha.

Reflection 17 (and a shout out to keenan)

To start, I dont care that this is not math....Keenan I'ma miss you dude. Riverside aint gone be the same wit you gone. It was tight havin you in some classes while it lasted. So good luck in life brother.

Ok so now to the math. We worked on study guides this week and there was a lot of remembering needed to do some of the early chapter tests for the study guide. One thing I remeber how to do, and its from way back in chapter 1, is how to work the i chart.

-When you get a number like say, i^3224, its easy to solve for it. All you have to do is divide the exponent (3224) by 4. In this case that will give you 86.

-Now use the i chart (plug the number you got when you divided by 4 (86) into the i chart:

.25 = i
.5 = -1
.75 = -i
1or any whole number = 1

-So because 86 is a whole number, i^3224 = 1. Yay!



Okay so another thing that was a great deal of fun at the beginning of the year that I remember well is logs. I'm going to show you some log stuff.

-To put a log into exponetial form use this handy dandy formula:
logb(x)=a ---logarithmic form
b^a=x ---exponential form

-Guess what....here's an example with numbers instead of letters:

logx(4)=2
-Put into exponetial form:
x^2=4

Reflection 17

this week was mainly focused on review for the exam and the bridges. congrats to those who held over 20 lbs by the way. I basically got a fresh review of chapters 1 thru 9 pounding in my head right about now.
IDENTITIES
csc=1/sin x
sec= 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*-x)
tan x=cot (90*-x)
sec x=csc (90*-x)
cos x=sin (90*-x)
cot x=tan (90*-x)
csc x=sec (90*-x)
tan x=sinx/cosx
cot x=cosx/sinx

if i am correct these should be all the identities we need to know

My Last Reflection...

This week we just worked on our study guides and review for our exam we learned how to use the identities to simplify equations, graph trig functions, and solve trig functions all by using identities for them.

The most common used Identities

Pythagorean Relationships:
sin^2 x + cos^2 x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x

Reciprocal Relationships:
csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x

Example:
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

reflection #17

this week was spent mostly as an exam review, and the bridge projects... which were awesome

one of the things i retained from chapter one (after looking back at it) was finding an intersection. this is how you do it...

eg: y=2x2-8x+5

step 1. does it open up or down?

if its positive it opens up, and if its negative it opens down

step 2. how many x-intercepts does it have?

b2-4ac discriminate: positive-2x intercepts negative-no x intercept 0-1 x intercept

(-8)2 -4(2)(5)=24 so there are 2 x intercepts

step 3. find the intercepts

complete the square to find

2x2-8x+5

2x2-8x =-5

x2-4x+4=-5/2+4

(x-2)2=3/2 [shortcut]

x-2= [square root of] 3/2

x-2= [square root] 3/ [square root] 2 X [square root] 2/ [square root]2

x=2+/- [square root] 6/2

put in point form

(2+ [square root]6/2, 0) (2- [square root] 6/2, 0)

step 4. find the y-intercept

plug in y

2(0)2-8(0)+5

=5

(0,5)

step 5. find the axis of symmetry

x=-b/2a

=-(-8)/2(2)

=2

step 6. find the vertex

x=2

2(2)2-8(2)+5

=-3

vertex= (2, -3)

reflection 17

this week we learned identities. there are many different identities you can use to make the problem look like its smaller. but you always have to remember to do the identities first and not the algebra. the algebra comes second. here are some examples on identities:.....

1. simplify:

(1-sinx)(1+sinx)
1-sinx+sinx-sinx^2
=1-sin^2
= cos^2x s^2+c^2=1
c^2=1-s^2


2. prove:

cotA(1+tan^2A)/tanA=csc^2A

cotA(sec^2A)/tanA

=1/tan9sec^2A)/tanA

=(sec^2A/tanA)/(tanA/1) = sec^2A/tan^2A

=(1/cos^2A)/(sin^2A/cos^2A) = cos^2A/ cos^2Asin^2A = 1/sin^2A


= csc^2A


my biggest problem with these identities is that i can never remember the identities and i never know what to do with the algebra part. it all looks so easy when we do it in class, but when im on my on doing homework or the packets im clueless and i get so confused.

Reflection

This reflection is a recap of Identities.Here is a list of the identities.

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

remember you must always start with an identity. DO NOT START WITH ALGEBRA.

Example:
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


Question. What is the formula for the sum and product to the roots of an equation?

Reflection 17.

The thing we learned this week was identities. In identities you pretty much just keep canceling things out. The thing you have to remember in identities is first you find an identity then you do some type of math(factoring, sandwich, ect.) to make the problem smaller. In this you have to memorize all of the identites it's the only way you will be able to work the problems.

These are the identites you will use a lot:

Pythagorean Relationships:
sin^2 x + cos^2 x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x

Reciprocal Relationships:
csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x

Examples**

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.

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.

Identites can be really confusing but you have to take it slow and do one step at a time. And remember that you always find an identity before trying to work it with math!
-----------------------------------

I kinda forgot all the stuff about conics. Like hyperbolas and cirles and ellipses. If someone could explain all that to me again that would be nice :)

Reflection 17

The law of sines states that in any triangle, the ratio between each angle and the side opposite of that angle is the same for all angles and opposite sides.

Look at figure one. The ratio between side a and angle a1 is the same as the ratio between side b and angle b1. Likewise the ratio between side b and angle b1 is the same as the ratio between side c and angle c1.




Another words:

(length of side a)/(sin a1) = (length of side b)/(sin b1) = (length of side c)/(sin c1)

In order to figure out the sin of a1, we need to divide the triangle into two right triangles as shown in figure two.

sin a1 = opposite side/hypotenuse = side h/(side b)

Likewise

sin b1 = opposite side/hypotenuse = side h/(side a)

Problem:

In figure one, side b is 2 inches long, side a is 3 inches long and angle a1 is 35. What is angle b1?

Solution:

According to the law of sines

side a/sin a1 = side b/sin b1

Multiply both sides of the equation by sine b1

sin b1 * side a/sin a1 = side b

Multiply both sides of the equation by sine a1

sin b1 * side a = side b * sin a1

Divide both sides of the equation by side a

sin b1 = side b * sin a1/side a

In order to solve for sin a1, we have to solve for the length of side h in figure two first.

We have the length of b and the angle a1. Side b, the angle a1 and side h form a right triangle with side b being the hypotenuse.

In this right triangle, side h is opposite angle a1.

sin a1 = side h/side b

multiply both sides of the equation by side b

sin a1 * side b = side h

We know the angle a1 and we know the length of side b. Therefore we can solve for the length of side h.

sin 35o * 2 = side h

sin 35o =0.574

0.574 * 2 = 1.148.

Side h is 1.148 inches long.

Now we have the following data:

side h = 1.148 inches

side b = 2

side a = 3

angle a1 = 35o

We can plug this data into the law of sines and solve for b1.

sin b1/side b = sin a1/side a

sin b1 = side b * sin a1/side a

sin b1 = 2 * 0.574/3 = 0.383

sin b1 = 0.383

arcsin 0.383 = b1

b1 = 22.5o

Problem 2:

In figure two, side h = 1.148 inches, side a = 3 inches and side b = 2 inches. What is the angle between side a and side b?

Solution:

The angle between side a and side b is angle c1 in figure one.

this week we learned how to use the identities to help simplify equations, graph trig functions, and solve trig functions using identities. this week i understood how to graph trig functions the most.
here is examples:
y=3cos(piex-2)+1
1. amp=3
2.perieod=2pie/pie=2 p/4=2/4=1/2
1.0
2.1/2
3.1
4.3/2
5.2
3. phase shift=2
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(piex-2)+1
***then you graph your equation with your amp., phase shift, and period.
4. vertical shift=1
***then after you find your vertical shift you move your graph that many spaces, in this case you move the graph one up.

what i didn't understand this week was how you simplify equations by using the identities. i don't understand how you use the identities and how you solve them.

Reflection #17

Okay, this week went by really fast. I guess it is because we have christmas break coming up and exams next week, but i don't know. We did not really learn anything new. We are just working on study guides for review for our exam. My blog today is going to be on review for the exam if anyone needs to refresh their memory. I'm going to give examples of how to find the zeros of an equation, the trig chart, and some identities to help you on the exam.
_________________________________________________________________________

Identities:


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
_____________________________________________________________________

Triangle Formulas:


SOHCAHTOA
sin = opposite leg/ hypotenuse
cos = adjacent leg / hypotenuse
tan = opposite leg / adjacent leg


area of a right triangle = 1/2 (b)(h)
area of an isoceles triangle = bh
area of a non-right triangle = 1/2 (leg) (leg) (sin (angle between those 2 legs)

_____________________________________________________________________

Six trig functions:
sin: y/r
cos: x/r
tan: y/x
cot: x/y
csc: r/y
sec: r/x

Unit Circle:
90(pi/2): (0,1)
180(pi): (-1,0)
270(3pi/2): (0,-1)
360(2pi):(1,0)
_____________________________________________________________________

Now the only thing that i did not understand through chapters 1-9, are the identities. I know what they are, but i just do not know how to put them in context with the problem. Can someone please give me an example of one? And i also do not get how to apply the trig chart sometimes. It gets a little confusing, but hopefully i can study and pass my exam on wednesday. I need example problems.

Reflection #17

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.

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.
(:

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 :)

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

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

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

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

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

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

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

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

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

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.

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

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:

Find
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.

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?

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

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.

________________________________________________________________


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.
_______________________________________________________________________

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.

_______________________________________________________________________

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.

____________________________________________________________________

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.

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

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.

________________________________________________________________


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.
_________________________________________________________________

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
_________________________________________________________________

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.

_________________________________________________________________

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.

___________________________________________________________________

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.

_____________________________________________________________________


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.
______________________________________________________________________

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)

_____________________________________________________________________

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.
_______________________________________________________________________

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.

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.

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.

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

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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

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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
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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.

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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.