Prove that cos θ cos θ/2 – cos 3θ cos 9θ/2 = sin7θ sin4θ

Using transformation formula, we get,

2 cos A cos B = cos(A + B) + cos (A – B)

-2 sin A sin B = cos(A + B) – cos (A – B)

Multiplying and dividing the expression by 2.

$\therefore$ LHS $=\frac{1}{2}\left(2 \cos \theta \cos \frac{\theta}{2}-2 \cos 3 \theta \cos \frac{9 \theta}{2}\right)$

Applying transformation formula, we get,

LHS $=\frac{1}{2}\left(\cos \left(\theta+\frac{\theta}{2}\right)+\cos \left(\theta-\frac{\theta}{2}\right)-\left\{\cos \left(3 \theta+\frac{9 \theta}{2}\right)+\cos \left(3 \theta-\frac{9 \theta}{2}\right)\right\}\right)$

$\Rightarrow \mathrm{LHS}=\frac{\frac{1}{2}}{2}\left(\cos \frac{30}{2}+\cos \frac{-}{2}-\cos \left(\frac{130}{2}\right)-\cos \left(-\frac{30}{2}\right)\right)$

$\Rightarrow$ LHS $=\frac{1}{2}\left(\cos \frac{3 \theta}{2}+\cos \frac{\theta}{2}-\cos \frac{15 \theta}{2}-\cos \frac{3 \theta}{2}\right)\{\because \cos (-x)=\cos x\}$

$\Rightarrow$ LHS $=\frac{1}{2}\left(\cos \frac{\theta}{2}-\cos \frac{15 \theta}{2}\right)$

$\Rightarrow$ LHS $={ }^{\frac{1}{2}}\left(2 \sin \left(\frac{\frac{\theta}{2}+\frac{15 \theta}{2}}{2}\right) \sin \left(\frac{\frac{15 \theta}{2}-\frac{\theta}{2}}{2}\right)\right)$

$\Rightarrow \mathrm{LHS}=\frac{1}{2}\left(2 \sin \left(\frac{8 \theta}{2}\right) \sin \left(\frac{7 \theta}{2}\right)\right)$

$\Rightarrow$ LHS $=\frac{1}{2}\left(2 \sin \left(\frac{8 \theta}{2}\right) \sin \left(\frac{7 \theta}{2}\right)\right)$

$\therefore$ LHS $=\sin 4 \theta \sin \left(\frac{7 \theta}{2}\right)=$ RHS

Hence,

$\cos \theta \cos \frac{\theta}{2}-\cos 3 \theta \cos \frac{9 \theta}{2}=\sin 4 \theta \sin \left(\frac{7 \theta}{2}\right)$