Question

The value of $\cos \theta \cdot \cos \frac{\theta }{2}-\cos 3\theta \cdot \cos \frac{9\theta }{2}$ is equal to

• A. $\sin 4\theta \cdot \sin \frac{7\theta }{2}$
• B. $\sin 8\theta$
• C. $\sin 7\theta +\sin 8\theta$
• D. $\sin 7\theta \cdot \sin 8\theta$

Answer 1

Answer: (A)

We have, $\cos \theta \cdot \cos \frac{\theta }{2}-\cos 3\theta \cdot \cos \frac{9\theta }{2}$
$=\frac{1}{2}\left[2\cos \theta \cdot \cos \frac{\theta }{2}-2\cos 3\theta \cdot \cos \frac{9\theta }{2}\right]$
$-\frac{1}{2}[\cos \left(\theta +\frac{\theta }{2}\right)+\cos \left(\theta -\frac{\theta }{2}\right)-$
$\left\{\cos \left(3\theta +\frac{9\theta }{2}\right)+\cos \left(3\theta -\frac{9\theta }{2}\right)\right\}]$
$=\frac{1}{2}\left[\cos \frac{3\theta }{2}+\cos \frac{\theta }{2}-\left\{\cos \frac{15\theta }{2}+\cos \left(-\frac{3\theta }{2}\right)\right\}\right]$
$=\frac{1}{2}\left[\cos \frac{3\theta }{2}+\cos \frac{\theta }{2}-\cos \frac{15\theta }{2}-\cos \frac{3\theta }{2}\right]$
$=\frac{1}{2}\left[2\sin \frac{\theta +15\theta }{4}\cdot \sin \frac{15\theta -\theta }{4}\right]$
$=\sin 4\theta \cdot \sin \frac{7\theta }{2}$