Question

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

Answer 1

(i)

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

(ii)

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

(iii)

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

(iv)

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

(v)

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