Question

sin 11theta / 4 sintheta /4 + sin 7theta / 4 sin 3theta /4

Answer 1

Dear student
$$\begin{gathered} \sin \left(\frac{11\theta }{4}\right)\sin \left(\frac{\theta }{4}\right)+\sin \left(\frac{7\theta }{4}\right)\sin \left(\frac{3\theta }{4}\right) \\ =\frac{\cos \left({\displaystyle \frac{11\theta }{4}}-{\displaystyle \frac{\theta }{4}}\right)-\cos \left({\displaystyle \frac{11\theta }{4}}+{\displaystyle \frac{\theta }{4}}\right)}{2}+\frac{\cos \left({\displaystyle \frac{7\theta }{4}}-{\displaystyle \frac{3\theta }{4}}\right)-\cos \left({\displaystyle \frac{7\theta }{4}}+{\displaystyle \frac{3\theta }{4}}\right)}{2} \\ \left[\because \sin x\sin y=\frac{\cos (x-y)-\cos (x+y)}{2}\right] \\ =\frac{\cos \left({\displaystyle \frac{5\theta }{2}}\right)-\cos 3\theta }{2}+\frac{\cos \theta -\cos \left({\displaystyle \frac{5\theta }{2}}\right)}{2} \\ =\frac{\cos \theta -\cos 3\theta }{2} \\ =\sin \left(\frac{\theta +3\theta }{2}\right)\sin \left(\frac{3\theta -\theta }{2}\right)\left[\because \cos x-\cos y=2\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{y-x}{2}\right)\right] \\ =\sin 2\theta .\sin \theta \end{gathered}$$
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