11-1

This is how to convert from Polar to Rectangular.

There are two formulas to use when converting.

Here are the two formulas to use

• x= r cos(ø)
• y= r sin(ø)

Instead of (x,y) it is (r,ø).

Example 1:

Convert (3,45º) to rectangular.

1. x=3cos45º, so x=1.576
2. y=3sin45º, so y=2.553
3. The rectangular coordinates are (1.576, 2.553)

-Sameer

6-4

This week is a golden opportunity to review hyperbolas. I am sure some of you remember this, but others of you may not so here we go with the blogging and the teaching and what not.

Standard form!

x^2/a^2-y^2/b^2=1 OR -x^2/a^2+y^2/b^2=1

1. Major axis is bigger denominator (i.e. positive)

2. Minor axis is smaller denominator (i.e. negative)

3. The vertex is square root of largest denominator (in point form the answer goes in place of the major axis)

4. The other intercept is the square root of the smallest denominator (see above for point form)

Steps 5 and 6 are unnecessary for hyperbolas.

7. f^2=larger denominator+smaller denominator OR f^2=vertex^2+other value^2

8. Asymptotes

y=+/- square root of y denom/square root x denom x

At the end you graph it, but I won't be doing that.

EXAMPLE

-x^2/16+y^2/36=1

1. y is major

2. x is minor

3. square root of 36=6

4. square root of 16=4

5 and 6 are still unnecessary.

7. f^2=16+36

f=square root of 52 or +/-2 square root of 13 (0,2square root 13)(0,-2square root 13)

8. y=+/-6/4= +/-3/2

Don't forget to graph after.

--Sarah

11-3

11-3 Powers of Complex Numbers
This week I will be reviewing how to find powers of complex numbers. We use De Moivre's Theorem to find these roots. It's simpler to find powers of a complex number when the complex number is in polar form.
De Moivre's Theorem:
If z = r cis theta, then z^n = r^n cis n(theta).

*In this section, you will not draw Argand diagrams.*

Example 1: Evaluate (2 cis 45 degrees)^2.

• z^2 = 2^2 cis 2(45 degrees)


• z^2 = 4 cis 90 degrees

Example 2: z = 3 cis 10 degrees. Use De Moivre's Theorem to find z^3.

• z^3 = 3^3 cis 3(10 degrees)


• z^3 = 27 cis 30 degrees

Example 3: If z = 1 + c, find z^4.

• z = x + yi


• r = sq. root of x^2 + y^2


• r = +/- sq. root of 2


• theta = tan^-1(y/x)


• theta = 45 degrees, 225 degrees


• z = sq. root of 2 cis 45 degrees


• z = sq. root of 2 cos 45 degrees + sq. root of 2 sin 45 degrees(i)


• z = - sq. root of 2 cos 225 degrees


• z^4 = sq. root of 2^4 cos 4(45 degrees)


• = 4 cis 180 degrees


• = r cos 180 degrees + 4 sin 180i

11-2

11-2 has problems that use complex numbers.
Formulas you need to know:

• z=x+yi
• z=rcos(theta)+rsin(theta)i
• z=rcis(theta)
• z=square root of x^2+y^2

When multiplying complex numbers:

• If it is a rectangular problem, you always use foil.
• If it is a polar problem, you always multiply r and add theta.

Example 1: 5+2i

• First, you have know if it is in rectangular form or polar. It is in rectangular form.
• Now, since you know it is in rectangular form, you know you will have to find r and theta.
• Use the formulas from above ^^^^^. To find r, you have to use the r=square root of x^2+y^2 formula.
  
• Put the numbers in the formula and solve: R=square root of 5^2+2^2=square root of 25+4=square root of 29. (i always equals 1)
  
• Now, you have to find theta by using the formula Theta=tan^-1(y/x).
• Put the numbers in the formula and solve: Theta=tan^-1(2/5)=21.801 and then you have to put it in quadrant one and then find the other answer which is 201.801.
• You can now put your answers in their final form: square root of 29cis21.801 and -square root of 29cis201.801.

Example 2: 8cis240 degrees

• Like the problem above, you have to know which form this equation is in. It is in polar form.
• Since it is in polar form, you now know that you have to find x and y.
• You have to use the formulas x=rcos(theta) and y=rsin(theta).
• Since the degrees in the original problem is not between 0-90, we have to find a reference angle. 240-180=60.
• Now you can place the numbers in the formulas.
• X=8cos60=8(1/2)=4
• Y=8sin60=8(square root of 3/2)=6.928
• Your final answer is z=4+6.928

-Amber :)

11-2

Today I'm going to teach 11-2, which involves solving problems using complex numbers. This is a fairly easy section. There are a few formulas. They are:

z = x+yi (in rectangular)

z = rcos(theta)+rsini OR rcis(theta) (in polar)

z = square root of x^2+y^2

Remember that i = square root of -1 and cannot be simplified.

To multiply complex numbers you must: 1. foil (if expressing in rectangular) 2. multiply r and add theta (if expressing in polar)

Now I have examples to help work out these problems.

Example 1: Express in polar form.

3-4i

First we need to find r and theta. This is when the formulas from 11-1 come back.

r = square root of 3^2 + (-4)^2

r = +/- 5

theta = tan^-1((-4)/3)

theta = 307 degrees7'48" , 127 degrees7'48"

Now we have to write it out in polar form. To do this you have to find out which quadrant your original formula falls in. Then that's how you determine which r go with which theta. Since 3,-4 falls in the foruth quadrant so r=5 and theta=307 degrees7'48" goes together. Your answers are:

z=5cis307degrees7'48"

z=-5cis127degrees7'48"

Example 2: Express in rectangular form

6cis100 degrees

In order to do this you have to find x and y first. Once again, formulas from 11-1 are going to be used again.

x = 6cos100 degrees

x = -1.04

y = 6sin100 degrees

y = 5.909

Now you just write the answer like an formula.

z = -1.04 + 5.909

-danaaaa(:

11-2

This week I am going to teach you all chapter 11 section 2. The problems in this section use complex numbers. There are several formulas that you need to know so that you can work problems in this section.
The formulas are as follows:
• z = x + yi
• z = r cos theta + r sin theta i
• z = r cis theta
• z = square root of ( x^2 + y^2 )
Remember, when multiplying complex numbers:
• To multiply rectangular equations, use FOIL
• To multiply polar equations, multiply r and add theta
Now I will show you an example of how to do one of these problems.
Example: Change ( 3, 3 ) from rectangular to polar
• r = sq. root of ( 3^2 + 3^2 )
• r = +/- 3 sq. root (2)
• theta = tan^-1 (3/3)
• theta = 45 degrees, 225 degrees
Your final answer would be (3 sq. root of 2, 45 degrees) and (-3 sq. root of 2, 225 degrees)

-Braxton-

11-1

So this week I am going to do my blog on 11-1, which has to do with polar problems. These problems are really easyyyyy. Polar problems are always (r, theta) and rectangular problems are always (x,y). There are a few formulas for these types of problems. So I am going to give you these formulas right nowww.



x=rcos theta
y=rsin theta
r=square root of x^2 + y^2
theta=tan^-1(y/x)


So I guess I am going to work a few examples nowww.




EXAMPLE 1: convert (3,4) into polar form.

• r=square root 3^2 + 4^2=5


• theta=tan^-1(4/3)=53.130 and 233.130


• so your answers are (5, 53 degrees 7 minutes 48 seconds) and (-5, 233 degrees 7 minutes 48 seconds)

EXAMPLE 2: plot the points (-4, 10 degrees)

OKAY, well that is it for this weeek.
-Brad.

11-2

This week I am going to explain how to work problems in 11-2, which are problems using complex numbers. There are a few formulas in this section. These formulas are:

• z=x+yi
• z=rcos theta + rsin theta i
• z=rcis theta
• |z|=square root of x^2 + y^2

Note: To multiply complex numbers:

• With rectangular problems remember to always foil.
• With polar problems you will multiply r and add theta.

Now I am going to do a few examples to help you understand how to work these problems.

Example 1: -1 + i

• Since it is in rectangular form, you are going to find r and theta.
• r=square root of -1^2 + 1^2. Which equals square root of 2.
• theta=tan^-1(-1). Which equals 45.
• When you put that on the quadrants, you will get 135 and 315.
• So once you do that, your answers will be square root of 2 cis 135 degrees and -square root of 2 cis 315 degrees.

Example 2: 6 cis 100 degrees

• x=6 cos 100= -1.042
• y=6 sin 100= 5.909
• Your answer is going to be -1.042 + 5.909

Well, that is all for this week. BYEEE

--Halie

11-3

today im going to show you how to de moivre's theorum. the formula for this is [z^n=r^n cis n (theta)]. it's really easy bc all you have to do is plug in the exponet the directions give you and then evaluate it .



EX: z= 4 cis 15 use de moivre's theorum to find z^ 3
 first you plug in the exponent in the formula
z^3= 4^3 cis 4 15 (degrees)
z^3= 64 cis 60 (degrees)

11-3

Chapter 11-3

Soo today, we are going to learn all about De Moivre’s theorem! As I said in an earlier blog, chapter 11 consists of a few formulas to be followed. Most of them are really easy. (Especially in this section because there is only one that you must know J) It is pretty simple if you follow the theorem exactly how it is stated. There is one thing that you need to keep in mind throughout this section and that is:

**DO NOT DRAW ARGAND DIAGRAMS**

De Moivre's theorem states the 2=rcisø then z^n=r^n cis(nø)

So now for the emphasis example!

Evaluate (2sin45)^2

1. z^2=2^2cis2(45)
2. z^2=4cis90

Since I don't think I hit a 150 words yet, I'll do a problem like this but working backwards.

z=4cis20º (Use De Moivre's theorem to find z^3)

1. z^3=(4)^3cis(3(20))
2. z^3=64cis60

So basically if you know and learn De Moivre's theorem, you can work any of these problems.

Hope you learned something!

Carleyyyy :)

11-4

I wasn't at school for this section (or 11-3) so i decided to do my blog on this section to see if I'm doing this right... I'm only using a similar example from the notes to help guide me through it

Find the fourth root of 16.
 (In the end i understand how to do "k" and all that but im not 100% sure how to find "n" though i think its found by the closest square root. For example, for 8, "n" would be 3 because 9 is the closest perfect square and the square root of 9 is 3 but yeah unsure.)

Hope you can read it. I honestly think it would be harder to understand if i would have typed it instead...

Polar to Rectangular

Convert (3,75degrees) to rectangular.

Formulas: x=rcos(theta) y=rsin(theta)
x=3cos(75) y=3sin(75)
=-2 =7/2
Answer: (0.77,2.9)

:P

So hopefully this posts, because I've been trying since Saturday and every time I hit post it deletes everything instead of posting it. I'm gonna talk about converting rectangular to polar and vice versa.

Convert (9,30 degrees) to rectangular

To do this, use the equation

x=9cos30 y=9sin30

You should get

(9 square root of 3/2,9/2)

Now, let's go over rectangular to polar.

Convert (3,3)

Use this equation:

x=square root of (3^2+3^2)=square root of 18 or +/-3square root of 2

theta=tan inverse (3/3) This equals one which is on your trig chart as 45 degrees. You'd then draw your coordinate plane and find that the other positive angle is 225 degrees. You should have two answers that are set up like this.

(3 square root of 2, 45 degrees)

(-3 square root of 2, 225 degrees)

That's that(:

--Sarah

11-1

Convert (4,120degrees) to rectangular. Formulas: x=rcos(theta) y=rsin(theta)
x=4cos(120) y=4sin(120)
 =-2                =7/2
 Answer: (-2,7/2)

11-1

So, chapter 11 was pretty easy chapter *thumbs up*. Today I'm going to teach section 1 of the chapter. That is polar and rectangular coordinates. I will teach how to convert from polar to rectangular and vice versa. Rectangular coordinates are in the form of (x,y), while polar coordinates are in the form of (r,theta). In order to do the convserions, you need the following formulas.

polar to rectangular : x= rcos(theta) y= rsin(theta)

rectangular to polar : r= square root of x^2+y^2 theta= tan^-1(y/x)

Here comes the examples!

Example 1 : Give the polar coordinates for (0,12)

r= square root of 0^2+12^

r= +/-12

theta = tan^-1(12/0)

theta = 90 degrees, 270 degrees

(12, 90 degrees), (-12,270 degrees)

Example 2 : Give the rectangular coordinates for (-3, 90 degrees)

x= -3cos(90 degrees)

x= -3(0)

x= 0

y= -3sin(90 degrees)

y= -3(1)

y= -3

(0,-3)

11-1

This section is all about polar coordinates and plots. In this section I will teach you all how to convert from rectangular coordinates to polar coordinates, convert from polar coordinates to rectangular coordinates, and plot polar coordinates. Rectangular coordinates are in the form of (x, y). Polar coordinates are in the form of (r, theta).

Whenever you convert from polar coordinates to rectangular coordinates, you use these two formulas:
• x = r cos ( theta )
• y = r cos ( theta )

Whenever you convert from rectangular coordinates to polar coordinates, you use these two formulas:
• r = sq. root of ( x^2 + y^2 )
• theta = tan^ -1 ( y/x )

Now I will show you some examples so that you can see how to work a problem that involves polar coordinates.

Example: Convert (5, 0) to polar
• r = sq. root (5^2 + 0^2)
• r = sq. root (25)
• r = +/- 5
• theta = tan^-1 (0/5)
• theta = 0
• (5,0) and (-5,0)


-Braxton

Chapter 11

We learned some pretty basic stuff this past week.

Today, we are going to learn how to convert from Polar to Rectangular.

It is easy to do it the other way too, but we're going to save that for next week :)

To convert from polar to rectangular, you need to know two fomulars.

These formulas are:

• x= r cos(ø)
• y= r sin(ø)

In the chapter, instead of using (x,y), you will be using (r,ø) in place of it.

Now for the example:

Convert (2,30º) to rectangular.

1. The first thing you need to know is that 2 is your r and 30º is your ø.
2. x=2cos30º, so x=sqrt 3
3. y=2sin30º, so y=1
4. Your rectangular coordinates is (sqrt 3, 1)

AND THAT IS HOW YOU CONVERT FROM POLAR TO RECTANGULAR :)

CARLEYYYYY(:

11-1

This section is on polar and rectangular coordinates.
When you graph, you have to use (r, theta) instead of (x,y).

There are a couple of formulas you need to know for this section.
When you convert from polar to rectangular: x=rcostheta, y=rsintheta.
When you convert from rectangular to polar: r=square root of x^2+y^2, theta=tan^-1(y/x).

Example 1: Give the polar coordinate for (6,3)

1. First, you have to use the formulas I gave you above ^^^. R=square root of 6^2+3^2= square root of 36+9=square root of 45. Theta=tan^-1 (3/6)=tan^-1 (1/2)=26 degrees 33 minutes 54 seconds, 206 degrees 33 minutes 54 seconds.
2. Now you have to find out what quadrant the point (6,3) is in: It's in the first quadrant.
3. Since it is in the first quadrant, you now have to find out what answer from your theta would put you in the first quadrant: that would be 26 degrees 33 minutes 54 seconds.
4. Now you can have your final answer which is: (square root of 45, 26 degrees 33 minutes 54 seconds) and (-square root of 45, 206 degrees 33 minutes 54 seconds.

Example 2: Give the rectangular coordinates for (6, 30 degrees).

1. Giving the rectangular coordinates is easier than giving the polar coordinates and it is faster.
2. The first thing you need to know is that 6 is your R and 30 degrees is your Theta.
3. Now you use your formulas from above ^^^^^.
4. X=RcosTheta=6cos30 degrees=6(square root of 3/2)=6square root of 3/2=3 square root of 3.
5. Y=RsinTheta=6sin30 degrees=6(1/2)=6/2=3.
6. Final answer: (3 square root of 3, 3).

-Amber :)

11-1 Polar

So this week, I am going to explain how to work Polar Problems. At least I think thats what they are called. So now I am going to give you a few formulas.

To convert from polar to rectangular, you need to know two fomulars. These formulas are:

1. x=r cos theta
2. y=r sin theta

To convert from rectangular to polar, you need to also know two formulas. These formulas are:

1. r= square root of x^2 + y^2
2. theta= tan -1 (y/x)

NOTE: Always graph using the angle (r, theta) instead od (x,y). That is VERY important.

So, now I am going to work a few examples for you.

Example 1: Plot points (2, 40 degrees)

• Since, I can't really explain this in much detail. I am just going to show you a picture of what those points would look like once you plot them and that is going to be your answer.

Example 2: Give the rectangular coordinate for (1, 20 degrees)

• This is when you use the formulas I gave you above.
• 1 = r and 20 degrees = theta
• x=1 cos 20
• y=1 sin 20
• Once you plug those into your calculator, you will get (.940, .342). Which is going to be your answer.

Example 3: Graph r=sin theta

• All you do here is plug this formula into your calculator after making sure you set the correct settings.
• NOTE: Make sure you are in radians and polar mode when graphing these types of problems.
• Now, I am just going to show you what the graph is going to look like.

Well, that is it for this weeeeekk. YAYYY!!! Hope I helped :D

--Halie :)

11-1

11-1 Polar Coordinates and Graphs

• When dealing with polar graphs, instead of using the standard coordinatets (x,y), we use the polar coordinates (r,theta). We can convert between rectangular and polar coordinates.


• Converting from polar to rectangular: x=(r)(cos)(theta) y=(r)(sin)(theta)


• Converting from rectangular to polar: r=sq.root of (x^2 + y^2) theta = tan^-1(y/x)

Example 1: Give the polar coordinates for (5,0).

1. r=sq. root of (5^2 + 0^2)


2. r=+/-5


3. theta=tan^-1(0/5)


4. theta=0 *Now you must find where tan is 0 on the unit circle, which is at 0 and 180.


5. Now, looking at the point (5,0), you see it is at the 0 degree mark, so your polar coordinates are (5,0 degrees).

Example 2: Give the rectangular coordinates for (4, 120 degrees).

1. x=(4)cos(120 degrees) *Find a reference angle off 120 in order to get a value off the trig chart. You should get -60, which is -1/2.


2. x=4(-1/2)


3. x=-2


4. y=(1)sin(120 degrees) *Find a reference angle just as you did when finding x. You should get -60, which is (sq.root of 3/2).


5. y=4((sq.root of 3)/2)


6. Your rectangular coordinates are (-2,(sq.root of 3)/2).

11-2

today im teaching you how to convert from rectangular to polar form. in polar form your answer should look like this: (r cos theta) + (r sin theta i). this can be abbreviated as (r cis theta). its is really easy to convert from rectangular to polar form. the formula is (the square root of: x^2 + y^2) then yo do (theta)= (tan inverse of: y/x). once you do this then  your answer should be z=(r cis theta). this would be considered in polar form.



EX: -1 + i
 first you do (the square root of: -1^2 + 1^2)
 this gives +/- (square root of 2)
then you do (theta)= (tan inverse of: 1/-1)
 you get = (tan inverse of: 1) *in the negative* 2nd and 4th quadrant
= 135 (degrees) and 315 (degrees)
the original points are in the 2nd quadrant
answer is [(square root of 2) cis 135(degrees)]
               [ -(square root of 2) cis 315 (degrees)]

more of those formulas that i've been blogging about for three whole weeks now because i don't know what else to do this on

So I am going to go over some tangent formulas because I think I have done everything else. These formulas are used when you want to do addition and subtraction with tangent rather than sine or cosine. Here we go, no more stalling.

tan (alpha-beta)= tan alpha-tan beta/1+tan alpha tan beta

tan (alpha+beta)=tan alpha+tan beta/1-tan alpha tan beta

EXAMPLE TIIIIIIIIIIIIIIIIIIIIIIIIIME

Suppose tan alpha=1/3 and tan beta= 1/2

Find tan (alpha+Beta)

First you're going to pick a formula. Obviously since it's addition we're going to be using the addition formula.

1/3+1/2/1-(1/3)(1/2)

Find a common denominator and add the top, multiply and subtract the bottom. After doing that, you should have:

5/6 / 5/6

This equals 1 (:

The problem may also ask you to show that alpha=tan inverse 1/3 +beta=tan inverse 1/2=pi/4

Simply show that tan inverse(1)=45 degrees=pi/4 (:

==Sarah

10-2

This week I'm going to explain how to do problems from 10-2 since I haven't done 10-2 yet. 10-2 uses two formulas that involves tan. It isn't hard since both of the problems look exactly the same except for the signs changing.

Formulas:

• Tan(a+b)=Tana+Tanb/1-TanaTanb
• Tan(a-b)=Tana-Tanb/1+TanaTanb

******A good way to remember the formulas is that the top sign is the same as the sign in the parentheses and the bottom sign is the opposite of the sign in the parentheses.******

Ex. 1: Tana=5/4 and Tanb=3/7. Solve for Tan(a-b).

• First, you have to plug the Tana and Tanb that it gave you into the formula.
• Tan(a-b)=Tana-Tanb/1+TanaTanb=(5/4)-(3/7)/1+(5/4)(3/7)
• Ok, now you have to solve the equation.
  
• (5/4)-(3/7)/1+(5/4)(3/7)=(35/28)-(12/28)/1+(15/28)
• Ok, so on the top of the equation I multiplied 5/4 by 7/7 and 3/7 by 4/4 because you can't subtract fractions unless they have a common denominator. Then on the bottom, all I did was multiply straight across.
• Continuing with the equation: (35/28)-(12/28)/1+(15/28)=(23/28)/(43/28)
• Now, this time on the bottom I replaced 1 with 28/28 and just added across because you need common denominators to add fractions.
• Continuing with the equation: (23/28)/(43/28)=644/1204=23/43. Final answer is 23/43. For this step, I just multiplied 23/28 by the reciprocal of 43/28 which is 28/43.

If you had to use the other formula for this problem, you solve it the same way. The only that is different are the signs.


-Amber :)

10-2

Soo today im am going to show you how to do problems with tan(a+or -b). There is 2 formulas that go with this section so im going to show you them and then work a problem or two and that should give you enough information to know how this section works. If you was wondering how my weekned went it went great!!!! I went to work satuday then when i got off i went to camp and spent the night and went hunting this morning for a while. I sat in a tree for about 4 hours and got eaten alive by these giant misquitos! Yea i know!! exciting!!! Well anyway im going to get back to math. EWWWWW.

soo here are the only 2 formulas you will need! :
*tan(a+B)=tan a + tan B / 1 - (tan a)(tan B)
*tan(a-B)=tan a - tan B / 1 + (tan a)(tan B)
WEll there are your formulas soo get ready because here come the examples!!!!!

the first example: Tan A=2/3 Tan B=1/2
tan(a+B)= (2/3) + (1/2) / 1 - (2/3)(1/2)
you can use your calculator and just plug this problem in and press enter and BAM!!!!
you get 7/4

Here is the second example!: Tana=2 Tanb=-1/3
tan(a-B)=2 - (-1/3) / 1 + (2)(-1/3)
once again you use your calculator and BAMM!!!!
you get 7! haha

Well now that i have just blown you away by this amazing blog im going to bed! goodnight!

10-4

Back an earlier chapter with this section. I'm just going to simplify this equation real quick. It's really basic but it's enough to get an idea of what to do and which function to use.
Simplify: sin^2 - 1
Now, for this trig function you would use either sin^2x + cos^2x = 1, 1 + tan^2x = sec^2x, or 1 + cot^2x = csc^2x
Since their is only one function that has sin^2 and a 1 in it, you would use sin^2x + cos^2x = 1 to simplify.
So what you do now is basically make the pythagorean relationship equal to sin^2 - 1 and what your left with on the other side of the equal sign, is your simplified answer.
So subtract the cos^2x and the 1 and get -cos^x as a final answer.
~ Parrish Masters

Chapter 10 Review

Well, I already did a blog for each section of chapter 10 and we did not learn anything new this week, so I am going to do a review blog. Then I will show you a different type of problem.

Formulas:
1. cos (alpha +/- beta) = cos (alpha) cos (beta) -/+ sin (alpha) sin (beta)
2. sin (alpha +/- beta) = sin (alpha) cos (beta) +/- cos (alpha) sin (beta)
3. sin x + sin y = 2 sin ((x + y)/(2) cos ((x-y)/(2)
4. sin x - sin y = 2 cos ((x+y)/(2) sin ((x-y)/(2)
5. cos x + cos y = 2 cos ((x+y)/(2) cos ((x-y)/(2)
6. cos x - cos y = -2 sin ((x+y)/(2) sin ((x-y)/(2)
7. tan (alpha + beta)= ((tan alpha + tan beta)/(1- tan alpha tan beta))
8. tan (alpha – beta)= ((tan alpha – tan beta)/(1+ tan alpha tan beta))
9. sin 2a= (2 sin a) (cos a)
10. cos 2a= (cos^2 a) – (sin^2 a)
11. cos 2a= 1- 2 sin^2 a
12. cos 2a= 2 cos^2 a -1
13. tan 2a= ( 2 tan a) / ( 1- tan^2 a)
14. sin a/2= +/- sq. root of ((1 – cos a) / (2))
15. cos a/2= +/- sq. root of (( 1+ cos a) / (2))
16. tan a/2= +/- sq. root of (( 1- cos a) / ( 1+cos a))
17. tan a/2= +/- (( sin a) / ( 1+ cos a))
18. tan a/2= +/- (( 1- cos a) / (sin a))
(Remember to replace, not plug in)

Example: Simplify 4 sin pi/6 cos pi/6
• 2(2 sin pi/6 cos pi/6)
• 2(sin 2(pi/6))
• 2(sin (pi/3))
• 2(sq. root 3/2)
• Sq. root of 3

-Braxton

10-2!!!

Today I'm go to explain how to do chapter 10, section 2. This is the sum and difference for tangent. There's only two formulas for this secction. They're almost the same thing except the sign changes.

Formulas:

tan(alpha+beta) = tan(alpha) + tan(beta)/1-tan(alpha)tan(beta)

tan(alpha-beta) = tan(alpha) - tan(beta)/1+tan(alpha)tan(beta)

REMINDER: You do not plug in for formulas like the ones above, you replace.

Okay, time for some examples!!!!

Suppose tan alpha = 1/3 and tan beta = 1/2

Find tan(alpha+beta)

= (1/3 + 1/2)/(1-(1/3)(1/2))

=( 2/6+3/6)/(1-1/6)

= (5/6)/(5/6)

= 1

Suppose tan alpha = 4/3 and tan beta = -1/2

Find tan(alpha+beta)

= (4/3+(-1/2))/(1-4/3(-1/2)

=(8/6+(-3/6))/(1-(-4/6)

=(5/6)/(10/6)

=1/2

And we're done!!!

(:

Chapter 10!!!

Sooooo, since we haven't learned any new stuff over the past couple of days, i decided we should go back to the first section of chapter 10 to review. The first thing we're gonna review is the formulas! do you remember them? You should! because we just took a test with them!!!

Sooo, here they areee.

1. cos(alpha+/-beta)=cos alpha cos beta-/+sin alpha sin beta

2. sin (alpha+/-beta)=sin alpha cos beta+/-cos alpha sin beta

3. sin x+sin y=2sinx+y/2 cos x-y/2

4. sin x-sin y=2cosx+y/c sin x-y/2

5. cos x+cos y=2cosx+y/2cosx-y/2

6. cos x-cos y=-2sinx+y/2 sin x-y/2

NOW WHAT DO YOU THINK WE'RE GOING TO DO NEXT?!?!?!?!?!?

Yep! You guessed it! We're going to do an example!!!!!

(REMEMBER: in this chapter you REPLACE, NOT PLUG IN!)

Find the exact value of Sin 15
1. Since sin 15 isn't on the trig chart, you must use a formula that will make it.
2. So you will use Sin(45-30)
3. You will now get, Sin45 cos 30-cos 45 sin30
4. Using the trig chart, you get Sqrt 2/2 x sqrt 3/2 - Sqrt 2/2 x sin 1/2
5. Simplify. You get (sqrt 6-sqrt 2)/4

and there's your review for CHAPTER 10 SECTION 1!!!!!!

CAARRRLLLLEEEEEYYYYYY :)

This is last weeks blog Mrs. Robinson!!!

Since i just did a whole blog im not gonna waste any time on this one.
THis blog is going to be about the cos(x+- y) stuff we did and finding exact values of stuff and all that good fun stuff that i love doing so much.

ok so im going to do a problem where you have to find the exact value of sin x

Ex: Find the exact value of sin 75

i think this was on the test but i dont think i did it right lol

You set the problem up as sin(75)= sin(45 + 30)

= sin(45)cos(30) + cos(45)sin(30)
=(s.r. 2/2)(s.r.3/2) + (s.r. 2/2)(1/2) (s.r.= square root)
= (s.r.6/4) + (s.r.2/4)
=(s.r. 6 + s.r. 2) / 4


YAY!! we did it again!! goodbye

Season 2 Coming Soon

Hello Everyone! Im sorry about the absence of the 'Cory Teaches You How to Do Something Show". I am here to inform you that Season 2 will be coming very soon, but for now you will just have to settle for ordinary boring blogs until i pick my grades up. When i say grades, i mean the 0 F i have in Wallers class and the soon to be F i will have in this class. Yes, i actually have a 0 as a final grade in Wallers class right now. Anywho lets get started.

Since i have no clue what we learned recently, im going to do this blog on the Tan formulas.

Formula: tan(x + y)= tan x + tan y/ 1 - tanxtany

Example: Suppose tan x= 3/4 and tan y= 2/3. Find tan(x + y)

tan(x + Y)= 3/4 + 2/3 / 1-(3/4)(2/3)
= 17/12 / 1- (1/2)
=17/12 / (1/2)
= 17/6

yay!!! WE DID IT!! goodbye

10-2

So, this week I am going to explain how to work problems using formuals for tan(a + or - B). There are two formulas for these types of problems. They are both exactly the same except for the signs change. So now I am going to give you the formulas.

1. tan(a+B)=tan a + tan B / 1 - (tan a)(tan B)


2. tan(a-B)=tan a - tan B / 1 + (tan a)(tan B)

Now that you have both of the formulas. I am going to work a few examples to help you understand better.

Example 1: tan a=2/3 tan B=1/2

Plug into the formula tan(a+B)

• tan(a+B)= (2/3) + (1/2) / 1 - (2/3)(1/2)


• After you plug into the formula, you just simply solve the problem using basic math.


• Your answer is then going to be 7/4

Example 2: tan a=2 tan B=-1/3

Plug into the formula tan(a-B)

• tan(a-B)=2 - (-1/3) / 1 + (2)(-1/3)


• Once again, all you have to do is solve using basic math.


• Your answer is then going to be 7

I hope you now understand how to work problems using these formula.

--Halie! :D

10-1

10-1 Formulas for cos (a +/ B) and sin (a +/ B)

There are two main formulas you will use in the section. They are the Sum and Difference Formulas for cosine and sine.
-cos (a +/ B) = cos a cos B /+ sin a sin B
-sin (a +/B) = sin a cos B +/ cos a sin B

There are four other formulas for this section that aren't used as often as the main two. These formulas are used when Rewriting a Sum or Difference as a Product.
-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)

Here are some examples:
Example 1: Find the exact value of 75.

• cos (45 + 30)


• cos 45 cos 30 - sin 45 sin 30 *Now, you change from degrees to radians by using the trig chart.


• (sq. root of 2/2) (sq. root of 3/2) - (sq. root of 2/2) (1/2)


• (sq. root of 6 - sq. root of 2)/4

Example 2: Simplify the given expression, cos 105 cos 15 + sin 105 sin 15.

• cos (105 - 15)


• cos 90


• 0

Example 3: Prove that the equation, cos (pi + x) = -sin x.

• cos pi cos x - sin pi sin x


• -cos x - 0


• -cos x

10-4

today im going to show you how to do 10-4. 10-4 is very similar to 8-5. it is almost bascially the same exact thing. you have to factor until you get something set to zero and then figure out what angles you have to get. this will make way better sense in the example i will shbow below so lets get to tht there example.



EX: sin^2 x = sin x
 first subtract sin x

sin^2 x - sin x = 0
factor out a sin x

sin x(sin x - 1) = 0
set both of them  = to 0
sin x = 0     sin x - 1 = 0
take the inverse of sin
x= inverse of sin (0)       x= inverse of sin (1)

sin is 0 on the unit circle at 0, 180, and 360 degrees
sin is 1 at 90 degrees.   it asks for the problem in degrees so u cross out 360
 x=0,90,and 180 degrees

10-2

The sum and difference formula is to be explained only by using the formulas below:
tan (alpha+beta)= tan (alpha) + tan (beta)/ 1- tan (alpha) tan (beta).

the differnce formula for tangent is tan (alpha-beta)= tan (alpha) - tan (beta)/ 1+ tan (alpha) tan (beta).

The key to getting your solution is remembering to REPLACE. Here is an example.

Find tan (alpha + beta) and tan (alpha-beta)
Ex: tan (alpha)= 3/2 tan (beta)= 1/3

(3/2) + (1/3) / 1- (3/2) (1/3)
= (9/6) + (2/6) / 1- (3/6)
= (11/6) / (6/6)-(3/6)
= 11/6 / 3/6 = 11/3

(3/2)-(1/3) / 1+(3/2) (1/3)
= (9/6)-(2/6) / 1+ (3/2) (1/3)
=(7/6) / (6/6) + (3/6)
= 7/6 / 9/6 = 7/9

* slash (/) means divided by and represents a fraction

-Sameer

10-1

Today I'm going to do my blog on chapter 10 section 1. This section is on the sums and difference for sine and cosine. This invloves six different formulas. The formulas are:

cos (alpha +/- beta) = cos (alpha) cos (beta) -/+ sin (alpha) (sin beta)

sin (alpha +/- beta) = sin (alpha) c0s (beta) +/- c0s (alpha) sin (beta)

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)

Here are some examples!!!

Simplify: sin (65 degrees) cos( 25 degrees) + cos (65 degrees) sin (25 degrees)

= sin (65 degrees + 25 degrees)

= sin 90 degrees

= 1

Simplify: cos ( 180 degrees) cos (120 degrees) + sin (180 degrees) sin (120 degrees)

= cos (180 - 120)

=sin 60 degrees

= 1/2

Well. there you go!

-Dana (:

i almost forgot to do this.

Like I said last week, chapter ten is nothing but formula after formula. So this blog is going to be some more formulas and then an example, but you probably knew that already. That's what most blogs consist of. Anyway here we go. Let's talk about addition and subtraction formulas...with tangent. I don't care if you don't want to. This is like airport security, you do what I say. Anyway, here are those formulas.

tan(alpha+beta)=tan alpha+tan beta/1-tan alpha tan beta

tan(alpha-beta)=tan alpha-tan beta/1+tan alpha tan beta

EXAMPLE

Find the exact value of the given expression:

tan 2pi/3+tan pi/12/1-tan 2pi/3 tan pi/12

First plug it into the equation (:

tan (2pi/3 + pi/12)

Converting to degrees is a good idea.

tan (270 degrees + 15 degrees)=tan 285 degrees

REFERENCE ANGLE TIME.

tan 285-360=-tan(45 degrees)

LOOK AT YOUR TRIG CHART, BUT YOU REALLY SHOULDN'T LOOK BECAUSE YOU SHOULD KNOW IT AND IF YOU DON'T YOU SHOULD BE ASHAMED OF YOURSELF.

final answer= -1

THAT'S HOW YOU DO THAT AND EVERYONE DESERVES A COOKIE :D

-SSSSSSSSSSSSSAAAAAAAAAAAAAAARRRRRRRRRRRRAAAAAAAAAAAAAHHHHHHHHH

10-3!!!!!

Soo today im going to do my blog on chapter 10 section 3! When i did my blog on chapter 10 section 1 last week you were able to tell that there were a few formulas well they also have a few formulas to go with chapter 10 section 3. So tonight i will begin with showing you all the formulas for double and half angles.

Half angle formulas:
sin(a/2)=+/-square root of (1-cosa)/2
cos(a/2)=+/-square root of (1+cosa)/2
tan(a/2)=1-cosa/sina
tan(a/2)=sina/1+cosa

Double angle formulas:
sin2a=2sinacosa
cos2a=cos^2-sin^2a
cos2a=2cos^2a-1
cos2a=1-2sin^2a
tan2a=2tana/1-tan^2a

So now that i have showed you all the formulas you need for this section i will do an example or 2 and show you how to use the formulas.

2sin30cos30
This is the first problem i will be showing you how to do. Some where in all these formulas there is one that relates to this problem! :sin2a=2sinacosa. There it is! yay!!!!
so then you will replace"a" with 30 and will end up with sin2(30) and that will give you sin60!!!!!

2tan80/1-tan^2(80)
this problem is similar to the one before you just fine the formula and replace.
tan2a=2tana/1-tan^2-> tan2(8)=tan160 degrees.

10-1

Last week I taught you all some formulas for trig equations. Those were double and half angle formulas. This week I am going to teach you how solve trig equations by using addition and subtraction formulas. This is actually two different sections. One is for sine and cosine, while the other is for tangent.
These are the formulas:
1. cos (alpha +/- beta) = cos (alpha) cos (beta) -/+ sin (alpha) sin (beta)
2. sin (alpha +/- beta) = sin (alpha) cos (beta) +/- cos (alpha) sin (beta)
3. sin x + sin y = 2 sin ((x + y)/(2) cos ((x-y)/(2)
4. sin x - sin y = 2 cos ((x+y)/(2) sin ((x-y)/(2)
5. cos x + cos y = 2 cos ((x+y)/(2) cos ((x-y)/(2)
6. cos x - cos y = -2 sin ((x+y)/(2) sin ((x-y)/(2)
7. tan (alpha + beta)= ((tan alpha + tan beta)/(1- tan alpha tan beta))
8. tan (alpha – beta)= ((tan alpha – tan beta)/(1+ tan alpha tan beta))

Example:
Simplify: sin 75 degrees cos 15 degrees + cos 75 degrees sin 15 degrees
• =sin (75 degrees + 15 degrees)
• =sin (90 degrees)
• =1

-Braxton

More Chapter 10

Ready for some more blogssssss?

Like I said in my earlier blog, chapter 10 is strictly formulas. It is a very easy concept once you know them all because all you will have to do is replace things. Today i am going to focus on double angles. I find these easier to work with because there is less simplifying to be done. Why this picture is so huge and refuses to get smaller, I am unsure of.

I'm going to show you an example.

1. If sine A= (3/5), find Sin 2A...

1. Find the formula you need to use:

Sin 2A=2sinA cosA

2. Make a triangle to find out what cosA is:

Sin 2A=2(3/5)(4/5)

3. Simplify:

Sin 2A=24/25

And that is how you do a Double Angle problem. I hope you learned something new this week! thanks

Carleyyyy :)

10-3

10-3 is on Half angles and double angles.
There are a lot of formulas in this section.

Half angle formulas:

• sin(a/2)=+/-square root of (1-cosa)/2
• cos(a/2)=+/-square root of (1+cosa)/2
• tan(a/2)=1-cosa/sina
• tan(a/2)=sina/1+cosa

Double angle formulas:

• sin2a=2sinacosa
• cos2a=cos^2-sin^2a
• cos2a=2cos^2a-1
• cos2a=1-2sin^2a
• tan2a=2tana/1-tan^2a

Ex. 1: 2sin20cos20

1. The first thing you would do for this is replace this problem with the formula it equals to.
2. So the formula you would replace it with is sin2a.
3. So that would give you sin2(20)=sin40 degrees
   

Ex. 2: 2tan60/1-tan^260

1. Just like the example above ^^^^, you would replace this problem with the formula it equals to.
2. The formula you would replace it with tan2a.
3. That would give you tan2(60)=tan120 degrees.

That's pretty much it for this section.
-Amber :)

10-3

10-3 Double-Angle and Half-Angle Formulas

We can use ten different trigonomic formulas in this section:

1. sin 2(a) = 2 sin(a) cos(a)



2. cos 2(a) = cos^2(a) - sin^2(a)



3. cos 2(a) = 1 - 2 sin^2(a)



4. cos 2(a) = 2 cos^2(a) - 1



5. tan 2(a) = 2 tan(a)/1 - tan^2(a)



6. sin a/2 = +/- square root of (1 - cos(a)/2)



7. cos a/2 = +/- square root of (1 + cos(a)/2)



8. tan a/2 = +/- square root of (1 - cos(a)/1 + cos(a))



9. tan a/2 = sin(a)/1 + cos(a)



10. tan a/2 = 1 - cos(a)/sin(a)

Example 1: Simplify the given expression.

1) 2 cos^2 10 - 1

• cos 2(10)



• cos(20)

2) cos^2 4A - sin^2 4A

• cos 2(4A)



• cos (8A)

Example 2: Find the exact value of the given expression.

1) 1 - 2 sin^2(7pi/12)

• cos 2(105)



• cos(210)



• Now you must find a reference angle: 210 - 180 = 30



• square root of 3/2

Example 3: If sin A = 5/13, find sin 2A.

• sin 2A = 2 sin A cos A



• sin 2A = 2 (5/13) (12/13)



• (10/13) (12/13)



• 120/169

-Jordan Duhon

10-3 Double Angle and Half Angle Formulas

This week I am going to explain how to work problems using double and half angle formulas. There are ALOT of formulas for this so you better be ready. Okay, so now I am going to list all the formulas for you.

Double Angle Formulas:

Half Anlge Formulas:

So now that you have all the formulas, I am going to work a few examples for you.

Example 1: 2 cos^2 10 degrees - 1

• All you would do for this problem is replace it with the formula it equals from above.
• Your answer would be cos 20 degrees.

Example 2: cos^2 4A - sin^2 4A

• All you have to do is again replace with a formula.
• Your answer would be cos 8A.

Example 3: 2 tan 25 degrees / 1-tan^2 25 degrees

• Once you would replace, you would get tan 50 degrees.

Well, I hope you understand how to work problems using double and half angles now. Glad I could help. Now I have to go work this chapter test that I have no clue how to do. GOOD BYE.

--Halie!

10-2

today im gonna show u the suma nd differnce formula for tangent and how it works. the sum formula is
tan (alpha+beta)= tan (alpha) + tan (beta)/ 1- tan (alpha) tan (beta). the differnce formula for tangent is
tan (alpha-beta)= tan (alpha) - tan (beta)/ 1+ tan (alpha) tan (beta). remember to replace. i will show you one example of how this works down below. so here it goes. yay.


       give tan (alpha + beta) and tan (alpha-beta)
Ex: tan (alpha)= 2/3     tan (beta)= 1/2

(2/3) + (1/2) / 1- (2/3) (1/2)
 = (4/6) + (3/6) / 1- (2/6)
= (7/6) / (6/6)-(2/6)
= 7/6 / 4/6       = 7/4

(2/3)-(1/2) / 1+(2/3) (1/2)
= (4/6)-(3/6) / 1+ (2/3) (1/2)
=(1/6) / (6/6) + (2/6)
= 1/6 / 8/6      = 1/8

10-1

Formulas:

sin (alpha plus or minus beta)=sin(alpha)cos(beta) plus or minus cos(alpha)sin(beta)

cos(alpha plus or minus beta)= cos(alpha)cos(beta) minus or plus sin(alpha) sin(beta)

Example: Find the exact value of sin 75

sin(45+30)= sin(45)cos(30)+cos(45)sin(30)

Replace angles with values found on the trig chart.

Solve to find your answer.

-Sameer