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Could someone explain how to simplify $\dfrac{\sin(2x)}{2-2\cos^2(x)}$? I've had tried the power reduction identity but the result did not seem much more simple. Any help would be appreciated.

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  • $\begingroup$ I think you should try using the identities $\sin^2x+\cos^2x=1$ and $\sin 2x=2\sin x\cos x$. $\endgroup$ – Peter Woolfitt Jun 27 '14 at 4:17
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How else can you write $\sin 2x$?

$\sin 2x = 2\sin x \cos x$

How else can you write $1 - \cos^2 x$?

$1 - \cos^2 x = \sin^2 x$

Substitute those and simplify.

$\dfrac{\sin 2x}{2 - 2 \cos^2 x} = \dfrac{\not 2 \not\sin x \cos x}{\not 2 \sin^{\not 2} x} = \boxed{\cot x}$

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