# Homework Help: Trig Identities

1. Nov 18, 2008

### Draggu

1. The problem statement, all variables and given/known data
Prove the following identity:

cos2A/1 + sin2A = cotA - 1 / cotA + 1

2. Relevant equations

3. The attempt at a solution

I proved the right side, which eventually lead up to cosA - sinA / cosA + sinA

I have NO idea how to do the left side. I have wasted roughly 10-15 sheets of paper now trying to prove it, but it all leads up to nothing. I tried MANY different solutions...

2. Nov 18, 2008

### marcuss

did u use the double angle formulas of sin2u=2sinucosu and cos2u=1-2sin^2u?

3. Nov 18, 2008

### -Vitaly-

http://img141.imageshack.us/img141/8388/trigom8.jpg [Broken]

Last edited by a moderator: May 3, 2017
4. Nov 18, 2008

### Draggu

Vitaly, thanks for trying BUT I have a question about where sin^2a(cot^2 - 1) came from.. could you add me to MSN please? supaflygt@hotmail.com

5. Nov 18, 2008

### -Vitaly-

http://img139.imageshack.us/img139/9676/trigkw2.jpg [Broken]
no probs
Look carefully, it's sine squared, not sin(2A). My 2A are big

Last edited by a moderator: May 3, 2017
6. Nov 18, 2008

### Draggu

Yeah, I know. I'm curious how you got sin^2a/sin^2a + sin^2a in the equation (2nd step) :P

7. Nov 18, 2008

### -Vitaly-

Please use brackets where necessary. (sinA)^2/(sinA)^2=1 and as you must know if you multiply something by 1, it doesn't change anything. It's a very common trick in math, e.g. integrating logs, taking inverse of a matrix, etc..
So, I muultipled (cosA)^2 by 1. And (sinA)^2 is therefore a common factor.
It's like (5+7)=(5*7/7+7)=7(5/7+1)

8. Nov 18, 2008

### Draggu

Well, I tried it doing my own thing, and here is the result:
EDIT: it's not -1 or 0, but it must be something.

9. Nov 18, 2008

### Draggu

Solved it now, thanks for your help.

10. Nov 19, 2008

### icystrike

my friend they are not equal to negative one. however you can prove both sides are
(cos A-sin A)/(cos A+sin A)

11. Nov 19, 2008

### -Vitaly-

My God, seeing those divisions just scares me.
Revisit fractions please and formulas of (a+b)2, a2-b2

And consider this: (52-32)/(5+3)2 is NOT equal -1.
(52-32)/(5+3)2=(5-3)(5+3)/((5+3)(5+3)). Divide by a common factor of 5+3, leaving you with (5-3)/(5+3)

Learn (a+b)2=(a+b)(a+b)=a2+2ab+b2, a2-b2=(a-b)(a+b) NOW

12. Nov 19, 2008

### Draggu

I know they aren't equal to -1. I proved both sides were (cosa-sina)/(cosa+sina), though, which was why I said it was solved

13. Nov 20, 2008

### icystrike

He says so because of your amazing cancellation.