0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.