Question

$\frac{\sin 5x}{\sin x}$ is equal to
(a) $16{\cos }^{4}x-12{\cos }^{2}x+1$
(b) $16{\cos }^{4}x+12{\cos }^{2}x+1$
(c) $16{\cos }^{4}x-12{\cos }^{2}x-1$
(d) $16{\cos }^{4}x+12{\cos }^{2}x-1$

Answer 1

(a) $16{\cos }^{4}x-12{\cos }^{2}x+1$

$$\begin{gathered} \mathrm{To}\mathrm{find}:\frac{\sin 5x}{\sin x} \\ \mathrm{Now}, \\ \sin 5x=\sin \left(3x+2x\right) \\ =\sin 3x\cos 2x+\cos 3x\sin 2x \\ =\left(3\sin x-4{\sin }^{3}x\right)\left(1-2{\sin }^{2}x\right)+\left(4{\cos }^{3}x-3\cos x\right)\left(2\sin x\cos x\right) \\ =\left(3\sin x-6{\sin }^{3}x-4{\sin }^{3}x+8{\sin }^{5}x\right)+2\sin x{\cos }^{2}x\left(4{\cos }^{2}x-3\right) \\ =\left(3\sin x-10{\sin }^{3}x+8{\sin }^{5}x\right)+2\sin x\left(1-{\sin }^{2}x\right)\left[4\left(1-{\sin }^{2}x\right)-3\right] \\ =\left(3\sin x-10{\sin }^{3}x+8{\sin }^{5}x\right)+\left(2\sin x-2{\sin }^{3}x\right)\left(4-4{\sin }^{2}x-3\right) \\ =\left(3\sin x-10{\sin }^{3}x+8{\sin }^{5}x\right)+\left(2\sin x-8{\sin }^{3}x-2{\sin }^{3}x+8{\sin }^{5}x\right) \\ =5\sin x-20{\sin }^{3}x+16{\sin }^{5}x \\ \therefore \frac{\sin 5x}{\sin x}=\frac{5\sin x-20{\sin }^{3}x+16{\sin }^{5}x}{\sin x} \\ =5-20{\sin }^{2}x+16{\sin }^{4}x \\ =5-20\left(1-{\cos }^{2}x\right)+16{\left(1-{\cos }^{2}x\right)}^2 \\ =5-20+20{\cos }^{2}x+16\left(1+{\cos }^{4}x-2{\cos }^{2}x\right) \\ =5-20+20{\cos }^{2}x+16+16{\cos }^{4}x-32{\cos }^{2}x \\ =16{\cos }^{4}x-12{\cos }^{2}x+1 \end{gathered}$$