Question

$\frac{\sin 5\mathrm{\theta }}{\sin \mathrm{\theta }}=$

(a) $16{\cos }^4\mathrm{\theta }-12{\cos }^2\mathrm{\theta }+1$
(b) $16{\cos }^4\mathrm{\theta }+12{\cos }^2\mathrm{\theta }+1$
(c) $16{\cos }^4\mathrm{\theta }-12{\cos }^2\mathrm{\theta }-1$
(d) $16{\cos }^4\mathrm{\theta }+12{\cos }^2\mathrm{\theta }-1$

Answer 1

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

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