Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
$\cos(x)$ is an even function. –  B. S. Dec 16 '12 at 19:06
    
Of course, thanks. –  revok Dec 16 '12 at 19:13

2 Answers 2

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.

share|improve this answer
    
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. :-) –  B. S. Dec 18 '12 at 13:39

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

share|improve this answer
    
${{+1^{+}}^+}^+ \quad \ddot\smile$ –  amWhy Apr 13 '13 at 0:31
    
@amWhy: Misssssss youuuuuu. –  B. S. Apr 17 '13 at 6:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.