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I would like to solve the following using only trig identities.

$$ \lim_{x \to \pi} {\cot2(x-\pi)}{\cot(x-\frac\pi2)} $$

I have so far that the above is equal to $$ \lim_{x \to \pi} \frac{\cos2(x-\pi)}{\sin2(x-\pi)}\frac{\cos(x-\frac\pi2)}{\sin(x-\frac\pi2)} = \lim_{x \to \pi} \frac{-\cos2x}{-\sin2x}\frac{-\sin x}{-\cos x} = \lim_{x \to \pi} \frac{\cos2x}{2\sin x\cos x}\frac{\sin x}{\cos x} = \frac12\lim_{x \to \pi} \frac{\cos2x}{\cos x}\frac{1}{\cos x}=\frac12 $$

However, I am afraid the correct answer is $-\frac12$. Where am I going wrong? Also, if there is another method of solving this, I would be thankful for the insight.

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$\cos(x)$ is an even function. – S. Snape Dec 16 '12 at 19:06
    
Of course, thanks. – revok Dec 16 '12 at 19:13
up vote 1 down vote accepted

First of all , $\sin(2(x-\pi)) = \sin(2x)$ and $\cos(2(x-\pi) )= \cos(2x)$. Look at the second step in your displayed equality. You have $\cos(x - \pi/2) = \cos(\pi/2 - x) = \sin(x)$, but $\sin(x - \pi/2) = -\sin(\pi/2 - x) = \cos(x).$ Apply these and you will get it.

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Thanks, I see that I didn't distribute the 2.. – revok Dec 16 '12 at 19:16
    
@revok: It's been good to up vote this answer. :-) – S. Snape Dec 18 '12 at 13:39

And $\cos(2x-2\pi)=\cos(2x)$ and $\cos(x-\frac{\pi}{2})=\sin(x)$ there.

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${{+1^{+}}^+}^+ \quad \ddot\smile$ – amWhy Apr 13 '13 at 0:31
    
@amWhy: Misssssss youuuuuu. – S. Snape Apr 17 '13 at 6:48

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