Question

Show that: $\cos 2\theta \cos \frac{\theta }{2}-\cos 3\theta \cos \frac{9\theta }{2}=\sin 5\theta \sin \frac{5\theta }{2}$.

Answer 1

$LHS$

$=\cos 2\theta \cos \frac{\theta }{2}-\cos 3\theta \cos \frac{9\theta }{2}$

$\frac{1}{2}[2\cos 2\theta \cos \frac{\theta }{2}-2\cos 3\theta \cos \frac{9\theta }{2}]$

(multiply & divide equation by $2$)

[now, we know that $2\cos A\cos B=\cos (A+B)+\cos (A-B)$]

$=\frac{1}{2}[\cos \left(\frac{5\theta }{2}\right)+\cos \left(\frac{3\theta }{2}\right)-\cos \left(\frac{15\theta }{2}\right)-\cos \left(\frac{-3\theta }{2}\right)]$

[now, $\cos$ Function is even i.e., $\cos (-\theta )=\cos \theta$]

so, $LHS=\frac{1}{2}[\cos \left(\frac{5\theta }{2}\right)-\cos \left(\frac{15\theta }{2}\right)]$

$=\frac{1}{2}\left[2\sin \left(\frac{5\theta +15\theta }{2\times 2}\right)\sin \left(\frac{15\theta -5\theta }{2\times 2}\right)\right]$

$(weknowthat,\cos A-\cos B=2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{B-A}{2}\right))$

$=\sin 5\theta \sin \frac{5\theta }{2}$

$=RHS$