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$$
\text { Prove that } \cos \theta \cos \frac{\theta}{2}-\cos 3 \theta \cos \frac{9 \theta}{2}=\sin 7 \theta \sin 8 \theta \text {. }
$$
\text { Prove that } \cos \theta \cos \frac{\theta}{2}-\cos 3 \theta \cos \frac{9 \theta}{2}=\sin 7 \theta \sin 8 \theta \text {. }
$$
Solution:
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Verified Answer
Since LHS $=\cos \theta \cos \frac{\theta}{2}-\cos 3 \theta \cos \frac{9 \theta}{2}$
$$
=\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)-\cos \left(3 \theta+\frac{9 \theta}{2}\right)\right.
$$
$\left.-\cos \left(3 \theta-\frac{9 \theta}{2}\right)\right]$
$$
=\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[\cos \frac{\theta}{2}-\cos \frac{15 \theta}{2}\right]
$$
$$
\begin{aligned}
&=-\frac{1}{2}\left[2 \sin \left(\frac{\theta+15 \theta}{2}\right) \cdot \sin \left(\frac{\theta-15 \theta}{2}\right)\right] \\
&{\left[\because \cos x-\cos y=-2 \sin \frac{x+y}{2} \cdot \sin \frac{x-y}{2}\right]} \\
&=+(\sin 8 \theta \cdot \sin 7 \theta)=\operatorname{RHS}
\end{aligned}
$$
$$
=\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)-\cos \left(3 \theta+\frac{9 \theta}{2}\right)\right.
$$
$\left.-\cos \left(3 \theta-\frac{9 \theta}{2}\right)\right]$
$$
=\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[\cos \frac{\theta}{2}-\cos \frac{15 \theta}{2}\right]
$$
$$
\begin{aligned}
&=-\frac{1}{2}\left[2 \sin \left(\frac{\theta+15 \theta}{2}\right) \cdot \sin \left(\frac{\theta-15 \theta}{2}\right)\right] \\
&{\left[\because \cos x-\cos y=-2 \sin \frac{x+y}{2} \cdot \sin \frac{x-y}{2}\right]} \\
&=+(\sin 8 \theta \cdot \sin 7 \theta)=\operatorname{RHS}
\end{aligned}
$$
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