Prove: $\cot\frac{x+y}{2}=-\left(\frac{\sin x-\sin y}{\cos x-\cos y}\right)$.

My original idea was to do this: $\cot\frac{x+y}{2}$ = $\frac{\cos\frac{x+y}{2}}{\sin\frac{x+y}{2}}$, then substitute in the formulas for $\cos\frac{x+y}{2}$ and $\sin\frac{x+y}{2}$, but that became messy very quickly.

Did I have the correct original idea, but overthink it, or is there any easier way? Hints only, please.

  • $\begingroup$ Do you know any formulas for $\sin x-\sin y$ and $\cos x-\cos y$? $\endgroup$ – CuriousGuest Jan 6 '15 at 23:29
  • $\begingroup$ @CuriousGuest Yes: $sin x - sin y = 2sin(\frac{x-y}{2})cos(\frac{x+y}{2})$ and $cos x - cos y = -2sin(\frac{x+y}{2})sin(\frac{x-y}{2})$ $\endgroup$ – Mathy Person Jan 6 '15 at 23:39

You should know the following factorisation formulae:

  • $\sin p +\sin q= 2\sin\dfrac{p+q}2 \cos\dfrac{p-q}2$

  • $\sin p-\sin q= 2\sin\dfrac{p-q}2 \cos\dfrac{p+q}2$

  • $\cos p +\cos q= 2\cos\dfrac{p+q}2 \cos\dfrac{p-q}2$

  • $\cos p-\cos q= -2\sin\dfrac{p+q}2 \sin\dfrac{p-q}2$

(They're derived from : \begin{align*} &\sin(a+b)+\sin(a-b)= 2\sin a \cos b && \sin(a+b)-\sin(a-b)= 2\sin b \cos a\\ &\cos(a+b)+\cos(a-b)= 2\cos a \cos b&& \cos(a+b)-\cos(a-b)= -2\sin a \sin b \end{align*} by setting $p=a+b$, $q=a-b$.)

  • $\begingroup$ $\cos p-\cos q= -2\sin\dfrac{p+q}2 \sin\dfrac{p-q}2$ $\endgroup$ – Narasimham Jan 7 '15 at 0:16
  • $\begingroup$ Oops! Thanks for pointing the typo! $\endgroup$ – Bernard Jan 7 '15 at 0:34

Try using

$ \displaystyle \sin{x} - \sin{y} = \sin{\frac{2x}{2}} - \sin{\frac{2y}{2}} = 2 \frac{\sin \left( {\frac{x+y}{2} + \frac{x-y}{2}} \right) - \sin \left( {\frac{x+y}{2} - \frac{x-y}{2}} \right)}{2}$

and then the product to sum-formula for sine, i.e.

$ \displaystyle \frac {\sin \left({x + y}\right) - \sin \left({x - y}\right)}{2} = \cos{x} \sin{y} $.

You can prove it geometrically as well.

  • $\begingroup$ I got: $\frac{sin(x)}{2} = \cos x \sin y$. Did I do that correctly? $\endgroup$ – Mathy Person Jan 7 '15 at 1:23
  • $\begingroup$ Not exactly. It is however correct that $\frac{\sin{2 x}}{2} = \sin{x} \cos(x)$. But in this case you want to utilize that $ \displaystyle \frac {\sin \left({u + v}\right) - \sin \left({u - v}\right)}{2} = \cos{u} \sin{v} $ setting $u = \frac{x+y}{2}$ and $v=\frac{x-y}{2}$. $\endgroup$ – Anton Jan 7 '15 at 9:11
  • $\begingroup$ $\ldots$ And then it should be as simple as using the definition of $\cot$. $\endgroup$ – Anton Jan 7 '15 at 9:18

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