Question

Prove that : $\frac{\cos 3\theta +\cos 3\varphi }{2\cos (0-\varphi )-1}$
$=(\cos \theta +\cos \varphi )\cos (0+\varphi )-(\sin \theta +\sin \varphi )\sin (\theta +\varphi )$

Answer 1

Take $D^r$ from L.H.S. to R.H.S.
R.H.S. = (cos $\theta$ + cos $4\pi$) [(cos $2\theta$ + cos 2$\varphi$) - cos($\theta$ + $\varphi$)]-(sin $\theta$ + sin $\varphi$)[(sin 2$\theta$ + sin $2\varphi$)-sin($\theta$ + $\varphi$)]
= $T_1+T_2$ each of which will have 6 terms on multiplication and then use
cos A cos B - sin A sin B = cos (A + B)
$\therefore$ R.H.S. = (cos $2\theta$ cos $\theta$ - sin $2\theta$ sin $\theta$) + (cos $2\varphi$ cos $2\varphi$ - sin $2\varphi$ sin $\varphi$) + cos($2\theta$ + $\varphi$) + cos($2\varphi$ + $\theta$) -cos($2\theta$ + $\varphi$) -cos($2\varphi$ + $\theta$)
= cos 3$\theta$ + cos 3$\varphi$, other terms cancel.