Question

Express the following as a sum or difference of two trigonometric ratios:

(i) $2\sin 4\theta \cdot \cos 2\theta$

(ii) $2\sin 2\theta \cdot \cos 4\theta$

(iii) $2\sin 2\theta \cdot \sin 4\theta$

(iv) $2\cos 2\theta \cdot \cos 4\theta$

Answer 1

Consider the equation,

Case 1- $2\sin 4\theta \cos 2\theta$

$\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

$=2\sin \left(\frac{6\theta +2\theta }{2}\right)\cos \left(\frac{6\theta -2\theta }{2}\right)$

$=2\sin 4\theta \cos 2\theta$

Case 2- $2\sin 2\theta \cos 4\theta$

$\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

$=2\cos \left(\frac{6\theta +2\theta }{2}\right)\sin \left(\frac{6\theta -2\theta }{2}\right)$

Case 3- $2\sin 2\theta \sin 4\theta$

$\cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$

$=-2\sin \left(\frac{6\theta +4\theta }{2}\right)\sin \left(\frac{6\theta -2\theta }{2}\right)$

$=2\sin 2\theta \sin 4\theta$

Case 4-$2\cos 2\theta \cos 4\theta$

$\cos A-\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

$=2\cos \left(\frac{6\theta +4\theta }{2}\right)\cos \left(\frac{6\theta -4\theta }{2}\right)$

$=2\cos 2\theta \cos 4\theta$

Hence, Proved.