How to solve this trig problem..

**acmair** · April 1st 2009, 08:50 PM

sinx +- siny / (cos x + cos y) = tan ((x+-y)/2)

Thank you!!

**Soroban** · April 2nd 2009, 06:24 AM

Hello, acmair!

Are we allowed the Sum-to-Product identities?

. . $\begin{array}{ccc}\sin A +\sin B &=& 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \\ \\[-3mm]<br /> \sin A - \sin B &=& 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \\ \\[-3mm]<br /> \cos A + \cos B &=& 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \\ \\[-3mm]<br /> \cos A - \cos B &=& \text{-}2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)<br /> \end{array}$

Prove: . $\frac{\sin x \pm \sin y}{\cos x + \cos y}\:=\:\tan \!\left(\frac{x\pm y}{2}\right)$

We have: . $\frac{\sin x + \sin y}{\cos x + \cos y} \;=\;\frac{2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})} {2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})} \;=\;\frac{\sin(\frac{x+y}{2})}{\cos(\frac{x+y}{2} )} \;=\;\tan\!\left(\frac{x+y}{2}\right)$

. . . And: . $\frac{\sin x - \sin y}{\cos x + \cos y} \;=\;\frac{2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2})} {2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})} \;=\;\frac{\sin(\frac{x-y}{2})} {\cos(\frac{x-y}{2})} \;=\;\tan\!\left(\frac{x-y}{2}\right)$

Therefore: . $\frac{\sin x \pm \sin y}{\cos x + \cos y} \;=\; \tan\!\left(\frac{x\pm y}{2}\right)$

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