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Math Help - How to solve this trig problem..

  1. #1
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    How to solve this trig problem..

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

    Thank you!!
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  2. #2
    Super Member

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    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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