# Limit of trigonometric functions using identities.

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. –  Babak S. Dec 16 '12 at 19:06
Of course, thanks. –  revok Dec 16 '12 at 19:13

## 2 Answers

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. :-) –  Babak S. 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. –  Babak S. Apr 17 '13 at 6:48