Express each of the following as the product of sines and cosines:

Solution:

(i) sin 12x + sin 4x

$=2 \sin \left(\frac{12 x+4 x}{2}\right) \cos \left(\frac{12 x-4 x}{2}\right)$   $\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \sin 8 x \cos 4 x$

(ii) sin 5x − sin x

$=2 \sin \left(\frac{5 x-x}{2}\right) \cos \left(\frac{5 x+x}{2}\right)$      $\left\{\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right\}$

 

$=2 \sin 2 x \cos 3 x$

(iii) cos 12x + cos 8x

$=2 \cos \left(\frac{12 x+8 x}{2}\right) c \operatorname{os}\left(\frac{12 x-8 x}{2}\right)$   $\left\{\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \cos 10 x \cos 2 x$

(iv) cos 12x − cos 4x

$=-2 \sin \left(\frac{12 x+4 x}{2}\right) \sin \left(\frac{12 x-4 x}{2}\right)$   $\left\{\because \cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\right\}$

$=-2 \sin 8 x \sin 4 x$

(v) sin 2x + cos 4x

$=\sin 2 x+\sin \left(\frac{\pi}{2}-4 x\right)$

$=2 \sin \left(\frac{2 x+\frac{\pi}{2}-4 x}{2}\right) \cos \left(\frac{2 x-\frac{\pi}{2}+4 x}{2}\right)$   $\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \sin \left(\frac{\pi}{4}-x\right) \cos \left(3 x-\frac{\pi}{4}\right)$