Question

Express each of the following as the product of sines and cosines:
(i) sin 12x + sin 4x
(ii) sin 5x − sin x
(iii) cos 12x + cos 8x
(iv) cos 12x − cos 4x
(v) sin 2x + cos 4x

Answer 1

(i)
$$\begin{gathered} \sin 12x+s\mathrm{in}4x \\ =2\sin \left(\frac{12x+4x}{2}\right)\cos \left(\frac{12x-4x}{2}\right)\left\{\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\sin 8x\cos 4x \end{gathered}$$

(ii)
$$\begin{gathered} \sin 5x-\sin x \\ =2\sin \left(\frac{5x-x}{2}\right)\cos \left(\frac{5x+x}{2}\right)\left\{\because \sin A-\sin B=2\sin \left(\frac{A-B}{2}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)\right\} \\ =2\sin 2x\cos 3x \end{gathered}$$

(iii)
$$\begin{gathered} \cos 12x+\cos 8x \\ =2\cos \left(\frac{12x+8x}{2}\right)c\mathrm{os}\left(\frac{12x-8x}{2}\right)\left\{\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\cos 10x\cos 2x \end{gathered}$$

(iv)
$$\begin{gathered} \cos 12x-\cos 4x \\ =-2\sin \left(\frac{12x+4x}{2}\right)\sin \left(\frac{12x-4x}{2}\right)\left\{\because \cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =-2\sin 8x\sin 4x \end{gathered}$$

(v)
$$\begin{gathered} \sin 2x+\cos 4x \\ =\sin 2x+\sin \left(\frac{\mathrm{\pi }}{2}-4x\right) \\ =2\sin \left(\frac{2x+{\displaystyle \frac{\mathrm{\pi }}{2}}-4x}{2}\right)\cos \left(\frac{2x-{\displaystyle \frac{\mathrm{\pi }}{2}}+4x}{2}\right)\left\{\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =2\sin \left(\frac{\mathrm{\pi }}{4}-x\right)\cos \left(3x-\frac{\mathrm{\pi }}{4}\right) \end{gathered}$$