Question

Solve

$\sin \frac{\theta }{2}\sin \frac{7\theta }{2}+\sin \frac{3\theta }{2}.\sin \frac{11\theta }{2}-\sin 2\theta .\sin 5\theta =$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Answer: (A)

$0$

$\sin \frac{\theta }{2}\sin \frac{7\theta }{2}=\frac{\cos (\frac{7\theta }{2}-\frac{\theta }{2})-\cos (\frac{7\theta }{2}+\frac{\theta }{2})}{2}$

$=\frac{\cos 3\theta -\cos 4\theta }{2}$

$sin\frac{3\theta }{2}\sin \frac{11\theta }{2}=\frac{\cos (\frac{11\theta }{2}-\frac{3\theta }{2})-\cos (\frac{11\theta }{2}+\frac{3\theta }{2})}{2}$

$=\frac{\cos 4\theta -\cos 7\theta }{2}$

$\sin 2\theta \cdot \sin 5\theta =\frac{\cos (\frac{5\theta }{2}-\frac{2\theta }{2})-\cos (\frac{5\theta }{2}+\frac{2\theta }{2})}{2}$

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

$\sin \frac{\theta }{2}\sin \frac{7\theta }{2}+sin\frac{3\theta }{2}\sin \frac{11\theta }{2}-\sin 2\theta \cdot \sin 5\theta$

$=\frac{\cos 3\theta -\cos 4\theta }{2}+\frac{\cos 4\theta -\cos 7\theta }{2}-\frac{\cos 3\theta -\cos 2\theta }{2}$

$=0$