Saturday, January 9, 2010
Reflection
This week went by slow, i guess it was because we just got back from the holidays. We started chapter 10 this week, so more stuff dealing with trig stuff and the trig chart. We took 2 quizzes, the first one we took i thought i did terrible on but did decent. Throughout the week, we ended up getting through the section 10-4. We learned a lot of formulas we have to memorize like: sum and differences with sin and cos, half and double angles, and many more. I found the easiest formula to work with was tan alpha + tan beta/1-tan alpha tan beta. I also found the sin(alpha + beta) was easy. But overall, this week was a simple week with mostly easy stuff we learned, just had to memorize a lot of formulas. And if anyone wants to help me with what we learned on friday, that would be great because i dont totally understand it.
Reflection #22
this week we learned how to use many formulas and the trig chart to help solve problems. here's the formulas that we learned:
Cos(α +/- β)=cos α cos β -/+ sin α sin β
sin(α +/- β)=sin α cos β -/+ cos α sin β
sin x + sin y= 2 sin x + y/2 cos x-y/2
sin x - sin y= 2 cos x + y/2 sin x-y/2
cos x + cos y= 2 cos x + y/2 cos x-y/2
cos x - cos y= 2 sin x + y/2 sin x-y/2
tan (α + β)=tan α + tan β/1-tan α tan β
tan (α - β)=tan α - tan β/1+tan α tan β
sin2α=2sin α cos α
cos 2α=cos^2 α –sin^2 α = 1-2 sin^2 α= 2 cos^2 α -1
tan 2α = 2tan α /1-tan^2 α
sin α/2= +/- √1-cos α/2
cos α/2= +/- √1+ cos α/2
tan α/2= +/- √1-cos α or 1 + cos α
=sin α/1+cos α
=1-cos α/sin α
something that i understood the most was section 2
here's some examples:
tan α = 2 and tan β=1
find tan (α - β)
= tan α + tan β/1-tan α tan β
=2+1/1-(2)(6)
=3/-1
=-3
Find the exact value of: tan 15+tan 30/1-tan 15 tan 30
tan α = 2 and tan β=1
find tan (α - β)
= tan (15 + 30)
=tan (45)
=1
for something that i didn't understand was in section 4 when you use the formulas as identities. i don't understand how you use the unit circle or the four quadrants to determine the degrees.
Cos(α +/- β)=cos α cos β -/+ sin α sin β
sin(α +/- β)=sin α cos β -/+ cos α sin β
sin x + sin y= 2 sin x + y/2 cos x-y/2
sin x - sin y= 2 cos x + y/2 sin x-y/2
cos x + cos y= 2 cos x + y/2 cos x-y/2
cos x - cos y= 2 sin x + y/2 sin x-y/2
tan (α + β)=tan α + tan β/1-tan α tan β
tan (α - β)=tan α - tan β/1+tan α tan β
sin2α=2sin α cos α
cos 2α=cos^2 α –sin^2 α = 1-2 sin^2 α= 2 cos^2 α -1
tan 2α = 2tan α /1-tan^2 α
sin α/2= +/- √1-cos α/2
cos α/2= +/- √1+ cos α/2
tan α/2= +/- √1-cos α or 1 + cos α
=sin α/1+cos α
=1-cos α/sin α
something that i understood the most was section 2
here's some examples:
tan α = 2 and tan β=1
find tan (α - β)
= tan α + tan β/1-tan α tan β
=2+1/1-(2)(6)
=3/-1
=-3
Find the exact value of: tan 15+tan 30/1-tan 15 tan 30
tan α = 2 and tan β=1
find tan (α - β)
= tan (15 + 30)
=tan (45)
=1
for something that i didn't understand was in section 4 when you use the formulas as identities. i don't understand how you use the unit circle or the four quadrants to determine the degrees.
Wednesday, January 6, 2010
Comments
tnagent formuals
tan(alpha+beta)=Tanalpha+tanbeta
1-tanalphatanbeta
tan(alpha-beta)=tanalpha-tanbeta
1+tanalphatanbeta
tan(alpha+beta)=Tanalpha+tanbeta
1-tanalphatanbeta
tan(alpha-beta)=tanalpha-tanbeta
1+tanalphatanbeta
Comments
Sum and Difference Formulas for Cosine and Sine
cos(alpha+beta)=cosalphacosbeta-+ sinalpha sinbeta
sin(alpha+-beta)=sinalphacosinebeat+- cosinealphasinebeta
cos(alpha+beta)=cosalphacosbeta-+ sinalpha sinbeta
sin(alpha+-beta)=sinalphacosinebeat+- cosinealphasinebeta
Late Reflection #21
im going to reveiw exponets. they can be pretty easy. example if your
given 16^1/4 it becoms 4square root 16 witch is equal to 2. heres
another example with exponets. 2^x=1/8 it then becomes
2^x=8^-1, 2^x=(2^3)^-1, 2^x=2^-3, so your answer will be x=-3.
something else ill review is logs. if your given logb x=a, so then you
take the number that the equation is equal to and rase it to the base and
set it all eaqual to the number that your given before the equal sign
witch is known as exponential form. that lookes like b^a=x.
you can also expand logs like log2 x y^2=log2 x + 2 log 2 y.
given 16^1/4 it becoms 4square root 16 witch is equal to 2. heres
another example with exponets. 2^x=1/8 it then becomes
2^x=8^-1, 2^x=(2^3)^-1, 2^x=2^-3, so your answer will be x=-3.
something else ill review is logs. if your given logb x=a, so then you
take the number that the equation is equal to and rase it to the base and
set it all eaqual to the number that your given before the equal sign
witch is known as exponential form. that lookes like b^a=x.
you can also expand logs like log2 x y^2=log2 x + 2 log 2 y.
Reflection
So this week we learned about how to add tan of alpha and beta to get an exact value.
The formula for adding tans of alpha and beta is: tan (alpah + beta) = tan alpha + tan beta over one minus tan alpha times tan beta.
Using this formula, you can find a numerical solution to a problem rather than a variable or measurement.
Tuesday, January 5, 2010
Late Reflection
So this week i'll do some reviewing with right triangles. I thought the right triangles were kind of simple and understandable, but when we moved on with stuff after that i started to not understand some things that i thought were harder.
EXAMPLES:
SOHCAHTOA
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
EXAMPLES:
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
EXAMPLES:
SOHCAHTOA
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
EXAMPLES:
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
Sunday, January 3, 2010
reflection 20
for trig inverses find 2 angles
first you need to determine which quadrants your angles are going to be in
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
first you need to determine which quadrants your angles are going to be in
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
reflection 19
Inequalities
3x-1>2
You add one to both sides...3x>3... and then you divide by three on both sides...x>1... and that is your answer. But you have to remeber if you multiply or divide by a negative number, then you have to switch the signs.....
3x-1>2
You add one to both sides...3x>3... and then you divide by three on both sides...x>1... and that is your answer. But you have to remeber if you multiply or divide by a negative number, then you have to switch the signs.....
For example
-2x+1>3
You would subtract one from both sides...-2x>2 and then you would divide by negative two...x<1...>
The absolute value ones were easy too. If you have on greater than or less than, it is an OR, but if you have one that is greater than or equal to or less than or equal to, it is an AND...........
-2x+1>3
You would subtract one from both sides...-2x>2 and then you would divide by negative two...x<1...>
The absolute value ones were easy too. If you have on greater than or less than, it is an OR, but if you have one that is greater than or equal to or less than or equal to, it is an AND...........
For example.
/x-4/ <5
So you make two problems:
x-4<5>-5 x<9>-1
Example of the AND problem:
/x-9/>or equal to 2
Instead of making two problems, you just add it to the front of the original problem:
-2>or=to x-9 >or=to 2
Then you add nine to all sides...7>or=to x >or=to 11
So your final answer would be xor=to 11
/x-4/ <5
So you make two problems:
x-4<5>-5 x<9>-1
Example of the AND problem:
/x-9/>or equal to 2
Instead of making two problems, you just add it to the front of the original problem:
-2>or=to x-9 >or=to 2
Then you add nine to all sides...7>or=to x >or=to 11
So your final answer would be x
relfection 18
Changing Bases
....use when a log cannot be solved
....used to solve for x as a variable
....used to change the base of log
1. exponetial form
2. take log of both sides
3. move exponents to the front
4. solve for variable
5. write as a function or whole number
solve:
log base 3 of 7
1. 3^x=7
2. log 3^x = log 7
3. x log 3= log 7
divide both sides by log 3 to get x by itself
x=log7/log3 would be you final answer.
solve
5^x=10
since the first step is already done for you move to step 2
log 5^x = log 10
x log 5 = 1
the reason for log 10 being 1 is bc the base of a log is understood to be 10
then divide to get x by itself and it would be: x=1/log 5
....use when a log cannot be solved
....used to solve for x as a variable
....used to change the base of log
1. exponetial form
2. take log of both sides
3. move exponents to the front
4. solve for variable
5. write as a function or whole number
solve:
log base 3 of 7
1. 3^x=7
2. log 3^x = log 7
3. x log 3= log 7
divide both sides by log 3 to get x by itself
x=log7/log3 would be you final answer.
solve
5^x=10
since the first step is already done for you move to step 2
log 5^x = log 10
x log 5 = 1
the reason for log 10 being 1 is bc the base of a log is understood to be 10
then divide to get x by itself and it would be: x=1/log 5
#20
hey everybody! its me again, bout to do another blog!!! haha
saved em both for tonight cuz i love them soo much! ha.....NOT!!!!!
Changing of Base:
used when a log cnt be solved
used to solve for x as a variable
used to change the base of a log
1.write as exponetial form
2. take the log base of both sides
3.move exponents to the front
4. solve for variable
5. write as a function or whole number if possible.
(IF ITS NOT POSSIBLE, LEAVE IN IN LOGARITHM FORM!)
_____________________________________________
AND AGAIN, no questions b/c of no school, PEACE!
saved em both for tonight cuz i love them soo much! ha.....NOT!!!!!
Changing of Base:
used when a log cnt be solved
used to solve for x as a variable
used to change the base of a log
1.write as exponetial form
2. take the log base of both sides
3.move exponents to the front
4. solve for variable
5. write as a function or whole number if possible.
(IF ITS NOT POSSIBLE, LEAVE IN IN LOGARITHM FORM!)
_____________________________________________
AND AGAIN, no questions b/c of no school, PEACE!
Reflection 20!
Oh, in the last blog i forgot to add that I DON'T WANNA GO BACK TOMORROW! Why can't we be normal and go back on Wednesday? REALLY? monday!
okay, wellll...
On the bright side, functions were pretty simple to me so i should probably explain them:
First, let's start with the fact that if you find the function of x, or f(x), you will find y.
Adding or subtracting functions:
(f+g)(x)
f(x)+g(x)
(f-g)(x)
f(x)-g(x)
example:
f(x)=x+1
g(x)=x-4
(f+g)(x)= x+1+x-4
(f+g)(x)=2x-3
Multiplying functions:
(fxg)(x)
f(x) x g(x)
f(x)=2x+4
g(x)=4x-6
(2x+4)(4x-6)
F.O.I.L.= (8x^2-12x+16x-24)
=8x^2+4x-24
Finding a function of another function:
(fog)(x)=
f(g(x))
f(x)= 2x+4
g(x)= x-4
plug in the g(x) equation into the f(x) equation where ALL x's appear.
2(x-4)+4
distribute.
2x-8+4
(fog)(x)=2x-12
--------------------------------------------------------
Unfortunately, i also forgot that i'm still quite iffy on domain and range.
thanksss!
okay, wellll...
On the bright side, functions were pretty simple to me so i should probably explain them:
First, let's start with the fact that if you find the function of x, or f(x), you will find y.
Adding or subtracting functions:
(f+g)(x)
f(x)+g(x)
(f-g)(x)
f(x)-g(x)
example:
f(x)=x+1
g(x)=x-4
(f+g)(x)= x+1+x-4
(f+g)(x)=2x-3
Multiplying functions:
(fxg)(x)
f(x) x g(x)
f(x)=2x+4
g(x)=4x-6
(2x+4)(4x-6)
F.O.I.L.= (8x^2-12x+16x-24)
=8x^2+4x-24
Finding a function of another function:
(fog)(x)=
f(g(x))
f(x)= 2x+4
g(x)= x-4
plug in the g(x) equation into the f(x) equation where ALL x's appear.
2(x-4)+4
distribute.
2x-8+4
(fog)(x)=2x-12
--------------------------------------------------------
Unfortunately, i also forgot that i'm still quite iffy on domain and range.
thanksss!
reflection 19
hey people, so math time again, school is dumb! :p
and this is comin straight from my math notebook sooo here it is:
the only time you don't find the range is when the problem is a fraction.
When working with a polynomial, the domain will always be (-infinity,infinity) but for range, it depends:
if the leading exponent is an odd number, the range is the same as the domain
if the leading exponent is 2, then it's [vertex,infinity)
if it's positive and if it's negative, it's (-infinity,vertex]
(The vertex is -b/2a).
For square roots:
1)set the inside = 0
2)set up a # line
3)try values on either side
4)eliminate anything -ve
5)set up intervals
Now, for fractions:
1)set bottom = 0
2)solve for x
3)set up interval
wish i would've known this for the exam! haha, i might've not have failed it!
______________________________________________________
no questions because i dont know what to ask because we didnt have school
and this is comin straight from my math notebook sooo here it is:
the only time you don't find the range is when the problem is a fraction.
When working with a polynomial, the domain will always be (-infinity,infinity) but for range, it depends:
if the leading exponent is an odd number, the range is the same as the domain
if the leading exponent is 2, then it's [vertex,infinity)
if it's positive and if it's negative, it's (-infinity,vertex]
(The vertex is -b/2a).
For square roots:
1)set the inside = 0
2)set up a # line
3)try values on either side
4)eliminate anything -ve
5)set up intervals
Now, for fractions:
1)set bottom = 0
2)solve for x
3)set up interval
wish i would've known this for the exam! haha, i might've not have failed it!
______________________________________________________
no questions because i dont know what to ask because we didnt have school
Reflection 19!
Sooo...
i chose to be stupid and do BOTH blogs on the same night.... whatta waste!
Anyways, i hope everyone had a happy & safe holidays, Christmas & New Year's was a blasttt! :)
On to the math, the easiest thing i've learned all year were exponential equations.
I'll try & explain these types of problems:
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
-------------------------------------------------------------------------------
As for now, i really don't have anything that i'm having EXTREME trouble with aside from SOHCAHTOA(
when /how to use it) so if anyone can try & refresh my memory, that'd be great! thanks!
i chose to be stupid and do BOTH blogs on the same night.... whatta waste!
Anyways, i hope everyone had a happy & safe holidays, Christmas & New Year's was a blasttt! :)
On to the math, the easiest thing i've learned all year were exponential equations.
I'll try & explain these types of problems:
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
-------------------------------------------------------------------------------
As for now, i really don't have anything that i'm having EXTREME trouble with aside from SOHCAHTOA(
when /how to use it) so if anyone can try & refresh my memory, that'd be great! thanks!
Reflection 20
So for my last reflection of the holidays I am blogging about trig inverses and the quadrants.
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
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
reflection 20
Last week of the holidays and back to school tomorrow, ew. Lets review law of sines, which is in chapter 9, i think 9-3 to be exact.. this formula may be useful in doing this: sinA/a=sinB/b=sinC/c. This formula is used when you know pairs in non right triangles. You are setting up a proportion, to make it easier. Heres some examples:....
1) An ABC triangle with b=123, a=16, and angleB= 115 degrees. Find A.
sin115/123=sinA/16
16sin115=123sinA
sinA=16sin115/123
A=sin-1((16sin115)/(123))
A= about 6.771 degrees
2) An ABC triangle with angleA=30 degrees, angleC=25 degrees, and a=4. Find c.
sin30/4=sin25/c
csin30=4sin25
c=4sin25/sin30
c= about 3.381 degrees
1) An ABC triangle with b=123, a=16, and angleB= 115 degrees. Find A.
sin115/123=sinA/16
16sin115=123sinA
sinA=16sin115/123
A=sin-1((16sin115)/(123))
A= about 6.771 degrees
2) An ABC triangle with angleA=30 degrees, angleC=25 degrees, and a=4. Find c.
sin30/4=sin25/c
csin30=4sin25
c=4sin25/sin30
c= about 3.381 degrees
reflection 21
Yea, we got school tomorrow, but i'm ready to get back, i enjoyed my holiday now i'm ready for the second half of the year lets bring it on!!!!!!!!!!! I'm gonna review a few things from every chapter we leanred before the holidays. 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 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. 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.
Reflection 20
Let's learn about some inequalities yall.
3x-1>2
You just add one to both sides...3x>3... and then you divide by three on both sides...x>1... and that is your answer. But you have to remeber if you multiply or divide by a negative number, then you have to switch the signs. For example,
-2x+1>3
You would subtract one from both sides...-2x>2 and then you would divide by negative two...x<1...>
The absolute zero ones were easy too. If you have on greater than or less than, it is an OR, but if you have one that is greater than or equal to or less than or equal to, it is an AND. For example.
(/) will be the absolute value sign
/x-4/ <5
So you make two problems:
x-4<5>-5 x<9>-1
Example of the AND problem:
/x-9/>or equal to 2
Instead of making two problems, you just add it to the front of the original problem:
-2>or=to x-9 >or=to 2
Then you add nine to all sides...7>or=to x >or=to 11
So your final answer would be xor=to 11
3x-1>2
You just add one to both sides...3x>3... and then you divide by three on both sides...x>1... and that is your answer. But you have to remeber if you multiply or divide by a negative number, then you have to switch the signs. For example,
-2x+1>3
You would subtract one from both sides...-2x>2 and then you would divide by negative two...x<1...>
The absolute zero ones were easy too. If you have on greater than or less than, it is an OR, but if you have one that is greater than or equal to or less than or equal to, it is an AND. For example.
(/) will be the absolute value sign
/x-4/ <5
So you make two problems:
x-4<5>-5 x<9>-1
Example of the AND problem:
/x-9/>or equal to 2
Instead of making two problems, you just add it to the front of the original problem:
-2>or=to x-9 >or=to 2
Then you add nine to all sides...7>or=to x >or=to 11
So your final answer would be x
Reflection 20
It's the end of another week, and it's time for another holiday reflection. Can't believe we're going back to school tomorrow, it's sad. Now, to the math.
Sine and Cosine curves
y= 3 cos(x)
Amplitude = how high the graph is
period = how long it takes for the graph to repeat
p/4 = what you add to each value to eventually get the period
phase shift = horizontal shift of the whole graph
vertical shift = up and down movement of the whole graph
y= 3 cos(x)
1. amp = 3
2. period = 2pi/b = 2pi/1 = 2 pi p/4 = 1/2 pi
values
1. 0
2. pi/2
3. pi
4. 3 pi/2
5. 2 pi
those are the final x values because there is no phase shift in the equation
we also learned about angles of inclination:
here are some formulas:
for a line
m = tan (alpha) where m=slope and alpha=angle of inclination
for a conic
tan 2(alpha) = B/A-C
for a conic if A = C then alpha = pi/4
Example: 2x + 5y = 15 find the angle of inclination
m= -2/5
tan(alpha)= -2/5
alpha = tan^-1 (-2/5)
alpha = 158.199 degrees and 338.199 degrees
To get this answer you have to use your trig inverse rules to get the two angles. these answers aren't exact, they are just approximate.
Sine and Cosine curves
y= 3 cos(x)
Amplitude = how high the graph is
period = how long it takes for the graph to repeat
p/4 = what you add to each value to eventually get the period
phase shift = horizontal shift of the whole graph
vertical shift = up and down movement of the whole graph
y= 3 cos(x)
1. amp = 3
2. period = 2pi/b = 2pi/1 = 2 pi p/4 = 1/2 pi
values
1. 0
2. pi/2
3. pi
4. 3 pi/2
5. 2 pi
those are the final x values because there is no phase shift in the equation
we also learned about angles of inclination:
here are some formulas:
for a line
m = tan (alpha) where m=slope and alpha=angle of inclination
for a conic
tan 2(alpha) = B/A-C
for a conic if A = C then alpha = pi/4
Example: 2x + 5y = 15 find the angle of inclination
m= -2/5
tan(alpha)= -2/5
alpha = tan^-1 (-2/5)
alpha = 158.199 degrees and 338.199 degrees
To get this answer you have to use your trig inverse rules to get the two angles. these answers aren't exact, they are just approximate.
Reflection 19
Some Log info...............
Changing Bases
-used when a log cnt be solved
-used to solve for x as a variable
-used to change the base of a log
1.write as exponetial form
2. take the log base of both sides
3.move exponents to the front
4. solve for variable
5. write as a function or whole number if possible.
*If not possibnle leave in log form...
solve:
log base 3 of 7
1. 3^x=7
2. log 3^x = log 7
3. x log 3= log 7
divide both sides by log 3 to get x by itself
x=log7/log3 would be you final answer.
solve
5^x=10
since the first step is already done for you move to step 2
log 5^x = log 10
x log 5 = 1
the reason for log 10 being 1 is bc the base of a log is understood to be 10
then divide to get x by itself and it would be: x=1/log 5
Changing Bases
-used when a log cnt be solved
-used to solve for x as a variable
-used to change the base of a log
1.write as exponetial form
2. take the log base of both sides
3.move exponents to the front
4. solve for variable
5. write as a function or whole number if possible.
*If not possibnle leave in log form...
solve:
log base 3 of 7
1. 3^x=7
2. log 3^x = log 7
3. x log 3= log 7
divide both sides by log 3 to get x by itself
x=log7/log3 would be you final answer.
solve
5^x=10
since the first step is already done for you move to step 2
log 5^x = log 10
x log 5 = 1
the reason for log 10 being 1 is bc the base of a log is understood to be 10
then divide to get x by itself and it would be: x=1/log 5
Reflection 18
EXAMPLES: Circles
Standard Form: (x-h)^2+(y-k)^2=r^2center: (h,k) radius: r
Find the center and radius of each cirlce.
1.) (x-3)^2+(y+7)^2=19center: (3,-7) radius: squareroot of 19
Find the intersection of the circle.
1.) x^2+4^2-25 and y=2x-2
a) y=2x-2
b) x^2=(2x-2)^2=25
c) x^2+4x^2-8x+4=25
5x^2-8x+4=25
5x^2-8x-21=0
5x^2-15x+7x-21=0
5x(x-3)+7(x-3)=0
(x-3)(5x+7)
x=3 x=-7/5
y=2(x)-2
2(3)-2=4
y=2(-7/5)-2=-24/5
Final Answer: (3,4) (-7/5,-24/5)
Write in Standard form.1.) Center: (4,3) Radius: 2(x-4)^2+(y-3)^2=4
Hyperbolas are still a little tricky...
Standard Form: (x-h)^2+(y-k)^2=r^2center: (h,k) radius: r
Find the center and radius of each cirlce.
1.) (x-3)^2+(y+7)^2=19center: (3,-7) radius: squareroot of 19
Find the intersection of the circle.
1.) x^2+4^2-25 and y=2x-2
a) y=2x-2
b) x^2=(2x-2)^2=25
c) x^2+4x^2-8x+4=25
5x^2-8x+4=25
5x^2-8x-21=0
5x^2-15x+7x-21=0
5x(x-3)+7(x-3)=0
(x-3)(5x+7)
x=3 x=-7/5
y=2(x)-2
2(3)-2=4
y=2(-7/5)-2=-24/5
Final Answer: (3,4) (-7/5,-24/5)
Write in Standard form.1.) Center: (4,3) Radius: 2(x-4)^2+(y-3)^2=4
Hyperbolas are still a little tricky...
REFLECTION!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
So flipping to a random page of my first notebook I find Right Triangles. This is dealing with trig. now.... SOH CAH TOA (Sine=opposite/hypotenuse)(Cosine =adjacent/hypotenuse)(Tangent=opposite/adjacent)
For csc, sec, and cot you do not have any special way to remember it so you just have to know it. csc=hypotenuse/opposite sec=hypotenuse/adjacent cot=adj/opp
now remember that SOH COH TOA can ONLY BE USED FOR RIGHT TRIANGLES. One angle must be 90 degrees.
remember also that sin deals with the y axis while cos deals with the x and of course tangent deals with both.
For csc, sec, and cot you do not have any special way to remember it so you just have to know it. csc=hypotenuse/opposite sec=hypotenuse/adjacent cot=adj/opp
now remember that SOH COH TOA can ONLY BE USED FOR RIGHT TRIANGLES. One angle must be 90 degrees.
remember also that sin deals with the y axis while cos deals with the x and of course tangent deals with both.
Reflection 20.
Alright, Once again I didn't bring my book home for the holidays so I'm winging this. The first thing that popped into my head was the thing where you convert degrees to radians, and radians to degrees. It's really easy you are either trying to get a number with pi or just a number. Or To convert degrees to radians you multiply the degrees by Pi/180. Your final answer will have Pi in it. To convert radians to degrees multiply by 180/Pi. Pi will cancel out and you will be left with just the number.
EXAMPLE: Convert to radians.
1.) 225 degrees
225 X pi/180
225/180 pi
45/36 pi
= 5/4 pi
2.) 30 degrees
30 X pi/180
30/180 pi
= 1/6 pi
Convert to degrees.
1.) 3/4 pi
3/4 pi X 180/pi
the pi cancels out
540/4
= 135 degrees
2.) pi/6
pi/6 X 180/pi
the pi cancels out
180/6
= 30 degrees
I'm not sure if that's enough words so something else I remember is coterminal angles. And you either add or subtract 360 to the number it gives you. If your looking for a negative coterminal angle you keep subtracting till you get a negative number. I think..I don't really remember too much so if it's wrong sorry:(
And that's pretty much it..
EXAMPLE: Convert to radians.
1.) 225 degrees
225 X pi/180
225/180 pi
45/36 pi
= 5/4 pi
2.) 30 degrees
30 X pi/180
30/180 pi
= 1/6 pi
Convert to degrees.
1.) 3/4 pi
3/4 pi X 180/pi
the pi cancels out
540/4
= 135 degrees
2.) pi/6
pi/6 X 180/pi
the pi cancels out
180/6
= 30 degrees
I'm not sure if that's enough words so something else I remember is coterminal angles. And you either add or subtract 360 to the number it gives you. If your looking for a negative coterminal angle you keep subtracting till you get a negative number. I think..I don't really remember too much so if it's wrong sorry:(
And that's pretty much it..
Reflection #20
For my second review blog ill talk about right triangles. All right triangles have a right angle of 90
degrees. To solve for any side of a right triangle you can use SOHCAHTOA. This means: sin =
opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent. And to solve for the
angles of the right triangle, you use the inverse of the trig functions and you can still use
SOHCAHTOA. Then to find the area of a right triangle you use this formula: A = 1/2(base)
(height). *Also, if you are given the lengths of two sides of the right triangle, you can use the
Pythagorean Theorem (c^2 = a^2 + b^2) to find the third side. Here’s an example: In triangle
ABC, angle A is 90 degrees, angle C is 30 degrees, side b is 12, and side c is 7. Find the length of side
a and the measure of angle B. Then find the area of the right triangle. So you use the Pythagorean
Theorem. So you get c^2 = 7^2 + 12^2. C^2 = 193. c = square root of 193. Then to find the
measure of angle B, all you have to do is add 90 degrees and 30 degrees and subtract the sum from
180.So you get 90 + 30 = 120. 180-120= 60. Angle B = 60 degrees. Now to find the area you just
plug your numbers into the formula >> A =1/2(base) (height). So you get A = 1/2(12) (7). Area = 42
degrees. To solve for any side of a right triangle you can use SOHCAHTOA. This means: sin =
opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent. And to solve for the
angles of the right triangle, you use the inverse of the trig functions and you can still use
SOHCAHTOA. Then to find the area of a right triangle you use this formula: A = 1/2(base)
(height). *Also, if you are given the lengths of two sides of the right triangle, you can use the
Pythagorean Theorem (c^2 = a^2 + b^2) to find the third side. Here’s an example: In triangle
ABC, angle A is 90 degrees, angle C is 30 degrees, side b is 12, and side c is 7. Find the length of side
a and the measure of angle B. Then find the area of the right triangle. So you use the Pythagorean
Theorem. So you get c^2 = 7^2 + 12^2. C^2 = 193. c = square root of 193. Then to find the
measure of angle B, all you have to do is add 90 degrees and 30 degrees and subtract the sum from
180.So you get 90 + 30 = 120. 180-120= 60. Angle B = 60 degrees. Now to find the area you just
plug your numbers into the formula >> A =1/2(base) (height). So you get A = 1/2(12) (7). Area = 42
Reflection #19
For my first review blog during the holidays I’m going to talk about logs there pretty simple, and
there’s many things that you can do with them. First you can solve example: if your given log2 4
in this problem your equation is always equal to x. so then you have log2 4=x you take the number
that is attached to the log and raise it to the variable like so x^2=4 and then you solve for x and the
answer would be x=2.
you can also condense them example: log 4 - log pi + 3 log r - log b remember that if there is no
base that it is 10. and when your condensing logs you can pull the log out of this you only need one
log, and remember -/divide, +/multiply. So now your answer will be: log(4r^3/pi b)
there’s many things that you can do with them. First you can solve example: if your given log2 4
in this problem your equation is always equal to x. so then you have log2 4=x you take the number
that is attached to the log and raise it to the variable like so x^2=4 and then you solve for x and the
answer would be x=2.
you can also condense them example: log 4 - log pi + 3 log r - log b remember that if there is no
base that it is 10. and when your condensing logs you can pull the log out of this you only need one
log, and remember -/divide, +/multiply. So now your answer will be: log(4r^3/pi b)
Reflection #20
Okay, I'm going to review synthetic division:
-when you divide a polynomial by a number
Ex. Divide p(x)=x^3+5x^2-2 by x+2
1st) solve x+2 for x...x=-2 (that's the number you're dividing by)
2nd) begin by listing each number (if there is not a term, replace it with 0)
Ex. -2) 1 5 0 -2
3rd) bring down the first number
ex. -2) 1 5 0 -2
...........1
4th) multiply 1 and -2 and place underneath second number
ex. -2) 1 5 0 -2
............-2
...........1
5th) add the second number and the number underneath (5 and -2)
ex. -2) 1 5 0 -2
............-2
...........1 3
6th) and repeat until done
ex. -2) 1 5 0 -2
............-2 -6 12
...........1 3 -6 10
7th) take the finished results and form an equation (the last number is the remainder)
ex. x^2+3x-6 R10
**if the remainder is 0, then the number you divided by is a factor
Ex. p(x)=2x^4+5x^3-8x^2-17x-6 divided by x-2
2) 2 5 -8 -17 -6
......4 18 20 6
...2 9 10 3 0
2x^3+9x^2+10x+3
-when you divide a polynomial by a number
Ex. Divide p(x)=x^3+5x^2-2 by x+2
1st) solve x+2 for x...x=-2 (that's the number you're dividing by)
2nd) begin by listing each number (if there is not a term, replace it with 0)
Ex. -2) 1 5 0 -2
3rd) bring down the first number
ex. -2) 1 5 0 -2
...........1
4th) multiply 1 and -2 and place underneath second number
ex. -2) 1 5 0 -2
............-2
...........1
5th) add the second number and the number underneath (5 and -2)
ex. -2) 1 5 0 -2
............-2
...........1 3
6th) and repeat until done
ex. -2) 1 5 0 -2
............-2 -6 12
...........1 3 -6 10
7th) take the finished results and form an equation (the last number is the remainder)
ex. x^2+3x-6 R10
**if the remainder is 0, then the number you divided by is a factor
Ex. p(x)=2x^4+5x^3-8x^2-17x-6 divided by x-2
2) 2 5 -8 -17 -6
......4 18 20 6
...2 9 10 3 0
2x^3+9x^2+10x+3
REFLECTION #20
Time to review again. yay. Umm let's see...I think I'll go over right triangles and isosceles triangles and whatnot. How to find the area, how the find an angle or side...yeah all that.
First I'll review right triangles. All right triangles have a right angle of 90 degrees. I'm sure we all know that by now. To solve for any side of a right triangle you can use SOHCAHTOA. This means: sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent. And to solve for the angles of the right triangle, you use the inverse of the trig functions and you can still use SOHCAHTOA. Then to find the area of a right triangle you use this formula: A = 1/2(base)(height). *Also, if you are given the lengths of two sides of the right triangle, you can use the pythagorean theorem (c^2 = a^2 + b^2) to find the third side.
Here are a few examples:
Ex. 1.) In triangle ABC, angle A is 90 degrees, angle C is 30 degrees, side b is 12, and side c is 7. Find the length of side a and the measure of angle B. Then find the area of the right triangle.
*This problem is really easy because all you have to do to find the length of the third side is use the pythagorean theorem.
*so you get c^2 = 7^2 + 12^2
*c^2 = 193
c = square root of 193
*Then to find the measure of angle B, all you have to do is add 90 degrees and 30 degrees and subtract the sum from 180.
*so you get 90 + 30 = 120
*180-120= 60
angle B = 60 degrees
*Now to find the area you just plug your numbers into the formula >> A = 1/2(base)(height)
*So you get A = 1/2(12)(7)
*Area = 42
Ex. 2.) In triangle ABC, angle A is 90 degrees, angle C is 45 degrees, and side c is 6. Find sides a and b, and angle B. Then find the area of the right triangle.
*Okay the easiest thing to do first is find angle B since we already know the measure of two other angles. So you just add 90 and 45, then subtract from 180.
*So angle B is 45 degrees.
*Now you can choose to solve for side a or b, it doesn't matter which one. I'll solve for side b first. Since you only have one side length, you can use SOHCAHTOA for this problem.
*So it would be tan45 = 6/b
*btan45 = 6 then divide by tan45
*Side b = 6
*Then to find the other side you use the pythagorean theorem.
c^2 = 6^2 + 6^2
c^2 = 72
*Side c = 6 square root of 2
*Now to find the area, you simply plug the numbers into the formula like this:
A = 1/2 (6)(6)
*Area = 18
Okay I think that's enough of right triangles for now. We learned how to solve isosceles triangles by using either law of sines or the law of cosines. You can only use the law of sines when you have a pair. (a corresponding angle and side) And you can use law of cosines for instance if you were given the length of all 3 sides and you were asked to find an angle. To find the area of a non-right triangle you have to use this formula: A = 1/2(leg)(leg)sin(angle in between)
Heres an example:
Ex. 3.) In triangle ABC, angle A is 35 degrees, side a is 3, and side b is 7. Solve the triangle for all missing sides and angles. Then find the area.
*Well first off, once you draw your picture, you see that you have a pair. So this means you can use the law of sines.
*So you get sin35/3 = sinB/7 ...then you cross multiply
7sin35 = 3sinB ..then divide by 3
sinB = 7sin35/3
*Since you are solving for an angle, you have to take the inverse of it like this:
B = sin^-1 (7sin35/3)
*angle B is approximately
....annnnndd i just messed up this problem. that's what i get for not bringing my notes home and making up this problem in my head. oh well. Is this good for a review? hope so.
Well bye, I'm going to macaroni grill now (:
First I'll review right triangles. All right triangles have a right angle of 90 degrees. I'm sure we all know that by now. To solve for any side of a right triangle you can use SOHCAHTOA. This means: sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent. And to solve for the angles of the right triangle, you use the inverse of the trig functions and you can still use SOHCAHTOA. Then to find the area of a right triangle you use this formula: A = 1/2(base)(height). *Also, if you are given the lengths of two sides of the right triangle, you can use the pythagorean theorem (c^2 = a^2 + b^2) to find the third side.
Here are a few examples:
Ex. 1.) In triangle ABC, angle A is 90 degrees, angle C is 30 degrees, side b is 12, and side c is 7. Find the length of side a and the measure of angle B. Then find the area of the right triangle.
*This problem is really easy because all you have to do to find the length of the third side is use the pythagorean theorem.
*so you get c^2 = 7^2 + 12^2
*c^2 = 193
c = square root of 193
*Then to find the measure of angle B, all you have to do is add 90 degrees and 30 degrees and subtract the sum from 180.
*so you get 90 + 30 = 120
*180-120= 60
angle B = 60 degrees
*Now to find the area you just plug your numbers into the formula >> A = 1/2(base)(height)
*So you get A = 1/2(12)(7)
*Area = 42
Ex. 2.) In triangle ABC, angle A is 90 degrees, angle C is 45 degrees, and side c is 6. Find sides a and b, and angle B. Then find the area of the right triangle.
*Okay the easiest thing to do first is find angle B since we already know the measure of two other angles. So you just add 90 and 45, then subtract from 180.
*So angle B is 45 degrees.
*Now you can choose to solve for side a or b, it doesn't matter which one. I'll solve for side b first. Since you only have one side length, you can use SOHCAHTOA for this problem.
*So it would be tan45 = 6/b
*btan45 = 6 then divide by tan45
*Side b = 6
*Then to find the other side you use the pythagorean theorem.
c^2 = 6^2 + 6^2
c^2 = 72
*Side c = 6 square root of 2
*Now to find the area, you simply plug the numbers into the formula like this:
A = 1/2 (6)(6)
*Area = 18
Okay I think that's enough of right triangles for now. We learned how to solve isosceles triangles by using either law of sines or the law of cosines. You can only use the law of sines when you have a pair. (a corresponding angle and side) And you can use law of cosines for instance if you were given the length of all 3 sides and you were asked to find an angle. To find the area of a non-right triangle you have to use this formula: A = 1/2(leg)(leg)sin(angle in between)
Heres an example:
Ex. 3.) In triangle ABC, angle A is 35 degrees, side a is 3, and side b is 7. Solve the triangle for all missing sides and angles. Then find the area.
*Well first off, once you draw your picture, you see that you have a pair. So this means you can use the law of sines.
*So you get sin35/3 = sinB/7 ...then you cross multiply
7sin35 = 3sinB ..then divide by 3
sinB = 7sin35/3
*Since you are solving for an angle, you have to take the inverse of it like this:
B = sin^-1 (7sin35/3)
*angle B is approximately
....annnnndd i just messed up this problem. that's what i get for not bringing my notes home and making up this problem in my head. oh well. Is this good for a review? hope so.
Well bye, I'm going to macaroni grill now (:
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