Proving Identity: cos2A/1 + sin2A = cotA - 1 / cotA + 1

[Draggu]

Homework Statement

Prove the following identity:

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

Homework Equations

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...

 

[marcuss]

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

 

[-Vitaly-]

http://img141.imageshack.us/img141/8388/trigom8.jpg

 

[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

 

[-Vitaly-]

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

 

[Draggu]

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

 

[-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)

 

[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.

 

[Draggu]

Solved it now, thanks for your help.

 

[icystrike]

Solved it now, thanks for your help.

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

 

[-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

 

[Draggu]

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

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

 

[icystrike]

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

He says so because of your amazing cancellation.