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  4. The value of cos θ ⋅ cos (θ/2)- cos 3 θ ⋅ cos (9 θ/2) is equal to

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

Trigonometric Functions

Solution:

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