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.

$$\frac{1}{\sin(z)} = \cot (z) + \tan (\tfrac{z}{2})$$

I did this:

First attempt: $$\displaystyle{\frac{1}{\sin (z)} = \frac{\cos (z)}{\sin (z)} + \frac{\sin (\frac{z}{2})}{ \cos (\frac{z}{2})} = \frac{\cos (z) }{\sin (z)} + \frac{2\sin(\frac{z}{4})\cos(\frac{z}{4})}{\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4})}} = $$ $$\frac{\cos (z)(\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4}))+2\sin z \sin(\frac{z}{4})\cos(\frac{z}{4})}{\sin (z)(\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4}))}$$

Stuck.

Second attempt:

$$\displaystyle{\frac{1}{\sin z} = \left(\frac{1}{2i}(e^{iz}-e^{-iz})\right)^{-1} = 2i\left(\frac{1}{e^{iz}-e^{-iz}}\right)}$$

Stuck.

Does anybody see a way to continue?

share|improve this question
1  
For your first strategy: You want to compare arguments $z$ and $z/2$, so you should halve $z$ with the formula, not $z/2$. For your second strategy: This is fine, now replace the right side also by exponentials and clear denominators. People post shorter solutions, but you should reflect that both of your ideas actually work fine. –  Phira Oct 31 '11 at 23:15

4 Answers 4

up vote 10 down vote accepted

$$ \frac{\cos (z)}{\sin (z)} + \frac{\sin (\frac{z}{2})}{ \cos (\frac{z}{2})} =\frac{\cos (z)\cos (\frac{z}{2})+ \sin(z)\sin (\frac{z}{2}) }{\sin (z)\cos (\frac{z}{2})} =\frac{\cos (z-\frac{z}{2})}{\sin (z)\cos (\frac{z}{2})}$$

share|improve this answer
2  
Rather elegant :-) –  joriki Oct 31 '11 at 22:35
    
$\displaystyle{ \frac{cos(-\frac{z}{2})}{sin(z) cos(\frac{z}{2}} = \frac{cos(\frac{z}{2})}{sin(z)cos(\frac{z}{2}} = \frac{1}{sin(z)} }$ –  VVV Oct 31 '11 at 22:44

Let $w = \frac{z}{2}$. Then $$ \cot(2w) + \tan(w) = \frac{\cos^2(w)-\sin^2(w)}{2 \sin(w) \cos(w)} + \frac{\sin(w)}{\cos(w)} = \frac{1}{\cos(w)} \left( \frac{\cos^2(w)-\sin^2(w) + 2 \sin^2(w)}{2 \sin(w)} \right) $$ The numerator becomes 1, and we arrive at the result $\frac{1}{2 \sin(w) \cos(w)} = \frac{1}{\cos(2w)} = \frac{1}{\cos(z)}$.

share|improve this answer

Start out with $$ \frac{1-\cos(z)}{\sin(z)}=\frac{2\sin^2(\tfrac{z}{2})}{2\sin(\tfrac{z}{2})\cos(\tfrac{z}{2})}=\tan(\tfrac{z}{2})\tag{1} $$ and add $\cot(z)$ to both sides: $$ \frac{1}{\sin(z)}=\cot(z)+\tan(\tfrac{z}{2})\tag{2} $$

share|improve this answer

I'll go backwards; I hope you don't mind.

$$\begin{align*}\cot\,z+\tan\frac{z}{2}&=\frac{\cos\,z}{\sin\,z}+\frac{\sin\,z}{1+\cos\,z}\\&=\frac{\sin^2 z+(1+\cos\,z)\cos\,z}{(1+\cos\,z)\sin\,z}\\&=\frac{\cos^2 z+\sin^2 z+\cos\,z}{(1+\cos\,z)\sin\,z}\\&=\frac{1+\cos\,z}{(1+\cos\,z)\sin\,z}\\&=\csc\,z\end{align*}$$

share|improve this answer

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.