# Verifying Identities [on hold]

How can we verify the following identity?

1. $-\frac{\sin 2x+\sin x}{\cos 2x-\cos x} = \cot(\frac{x}{2})$

These are the double angle identities that can be applied:

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

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

$\tan 2x = \frac{2\tan x}{1-\tan^2 x}$

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## put on hold as off-topic by Najib Idrissi, Adriano, drhab, Old John, Giuseppe Negroyesterday

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Whats the difference? –  Valtteri Feb 19 '13 at 21:43
I turned them into identities. They can now be verified. –  user62991 Feb 19 '13 at 21:46
What have you tried? What multiple angle identities do you know? They would be helpful here. –  Ross Millikan Feb 19 '13 at 21:46
@Valtteri Statements have a truth value, they're either false or true. Not all expressions are statements. –  Git Gud Feb 19 '13 at 21:47

Use $\sin2x=2\sin x\cos x$ and $\cos2x=2\cos^2x-1$ and cancel $\cos2x+1$ top and bottom to get it down to proving $${\sin x\over1-\cos x}=\cot(x/2)$$ Then use $\sin x=2\sin(x/2)\cos(x/2)$ and $\cos x=1-2\sin^2(x/2)$.

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HINT:

Using $$\sin C+\sin D=2\sin\frac{C+D}2\cos\frac{C-D}2$$

$$\text{and }\cos C-\cos D=-2\sin\frac{C+D}2\sin\frac{C-D}2,$$

$$\frac{\sin C+\sin D}{\cos C-\cos D}=-\cot\frac{C-D}2\text{ assuming } \sin\frac{C+D}2\ne0$$

Here $C=2x,D=x$

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