Proving trig equation

**michaelleung** · March 12th 2011, 05:19 PM

I'm having trouble proving trig identities. Especially when the answers don't seem too obvious (you know, cancel something out here, cancel something out there, and then there's the answer). Here's the question I'm stuck on:

$\frac{\sin2\theta}{2-2\cos^2\theta}$ = $\cot\theta$

If anyone can help, that would be appreciated!

**cuteangel** · March 12th 2011, 05:44 PM

$\frac{\sin2\theta}{2-2\cos^2\theta}$ = $\cot\theta$

In this equation, try replacing $\sin2\theta$ with its identity = $2\sin\theta\cos\theta$
In the denominator, try factoring out the 2 and seeing what the equation could be equal to.
Recall $\sin^2\theta + \cos^2\theta = 1$

**michaelleung** · March 12th 2011, 06:10 PM

Originally Posted by cuteangel

$\frac{\sin2\theta}{2-2\cos^2\theta}$ = $\cot\theta$

In this equation, try replacing $\sin2\theta$ with its identity = $2\sin\theta\cos\theta$
In the denominator, try factoring out the 2 and seeing what the equation could be equal to.
Recall $\sin^2\theta + \cos^2\theta = 1$

So I get $2-2\sin^2\theta$ on the bottom?

**cuteangel** · March 12th 2011, 06:15 PM

Well, not exactly.
$2 - 2\cos^2\theta = 2(1 - cos^2\theta)$
Now, rearranging $\sin^2\theta + \cos^2\theta = 1$ gives
$1 - \cos^2\theta = \sin^2\theta$
If you substitute this value in now, the equation should be solvable.

**Prove It** · March 12th 2011, 06:17 PM

Can you take out a common factor in the denominator? Can it be simplified using the Pythagorean Identity?

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