Prove that $\cos \theta \cos\large\frac{\theta}{2}-$$\cos 3\theta\cos \large\frac{9\theta}{2}=$$\sin 7\theta\sin 8\theta$

Toolbox:
• $\cos a\cos b=\large\frac{1}{2}$$[\cos(a+b)-\cos(a-b)] • \cos a-\cos b=-2\sin (\large\frac{a+b}{2})$$.\sin (\large\frac{a-b}{2})$
• $\sin (-\theta)=-\sin \theta$
• $\cos (-\theta)=\cos \theta$
LHS
$\cos \theta.\cos \large\frac{\theta}{2}$$-\cos 3\theta\cos \large\frac{9\theta}{2} \large\frac{1}{2}$$[\cos(\theta+\large\frac{\theta}{2})]-\large\frac{1}{2}$$[\cos(3\theta+\large\frac{9\theta}{2}]$$+\cos(3\theta-\large\frac{9\theta}{2})]$
$\large\frac{1}{2}$$[\cos \large\frac{3\theta}{2}+$$\cos\large\frac{\theta}{2}]-\large\frac{1}{2}$$[\cos \large\frac{15\theta}{2}+$$\cos\large\frac{-3\theta}{2}]$
$\large\frac{1}{2}$$[\cos \large\frac{3\theta}{2}$$+\cos \large\frac{\theta}{2}-$$\cos \large\frac{15\theta}{2}$$-\cos \large\frac{3\theta}{2}]$
$\large\frac{1}{2}$$[\cos \large\frac{\theta}{2}$$-\cos \large\frac{15\theta}{2}]$
$\large\frac{1}{2}$$\times -2\sin (\large\frac{\theta}{2}+\frac{15\theta}{2})$$\sin (\large\frac{\theta}{2}-\frac{15\theta}{2})$