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I'm currently trying to prove the following trig identity.

$\dfrac{\sin \left ( \frac{\alpha}{2} \right ) \cos \left ( \frac{\alpha}{2} \right ) + \sin \left ( \frac{\beta}{2} \right ) \cos \left ( \frac{\beta}{2} \right ) }{\cos ^2 \left ( \frac{\alpha}{2} \right ) - \sin^2 \left ( \frac{\beta}{2} \right )} = \dfrac{\sin \left ( \frac{\alpha+\beta}{2} \right )}{\cos \left ( \frac{\alpha+\beta}{2} \right )} $

Is there a way to do this without the use of sums to products/products to sums identities? It can easily be done using that, but I was wondering if there is another way out there.

Trogdor
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1 Answers1

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Use $(i)\sin2A=2\sin A\cos A$

and $(ii) $Prosthaphaeresis $\sin C+\sin D$

or if Prosthaphaeresis is prohibited, write $\alpha=\dfrac{\alpha+\beta}2+\dfrac{\alpha-\beta}2$ and $\beta=\dfrac{\alpha+\beta}2-\dfrac{\alpha-\beta}2$

and finally utilize $(iii)$ Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$

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lab bhattacharjee
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