# Proving trig equation

• Mar 12th 2011, 05:19 PM
michaelleung
Proving trig equation
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!
• Mar 12th 2011, 05:44 PM
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 $\sin^2\theta + \cos^2\theta = 1$
• Mar 12th 2011, 06:10 PM
michaelleung
Quote:

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?
• Mar 12th 2011, 06:15 PM
cuteangel
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.
• Mar 12th 2011, 06:17 PM
Prove It
Can you take out a common factor in the denominator? Can it be simplified using the Pythagorean Identity?