Question

The value of $\frac{\sin 5\theta }{\sin \theta }$ =

• A. $16{\cos }^4\theta -12{\cos }^2\theta +1$
• B. $16{\cos }^4\theta +12{\cos }^2\theta +1$
• C. $16{\cos }^4\theta -12{\cos }^2\theta -1$
• D. $16{\cos }^4\theta +12{\cos }^2\theta -1$

Answer 1

The correct option is A $16{\cos }^4\theta -12{\cos }^2\theta +1$

$\sin 5\theta =\sin (3\theta +2\theta )$

$⟹\sin 5\theta =\sin 3\theta \cos 2\theta +\cos 3\theta \sin 2\theta$

$⟹\sin 5\theta =(3\sin \theta -4{\sin }^3\theta )(2{\cos }^2\theta -1)+(4{\cos }^3\theta -3\cos \theta )\left(2\sin \theta \cos \theta \right)$

Therefore, $\frac{\sin 5\theta }{\sin \theta }=(3-4{\sin }^2\theta )(2{\cos }^2\theta -1)+(4{\cos }^3\theta -3\cos \theta )\left(2\cos \theta \right)$

$⟹\frac{\sin 5\theta }{\sin \theta }=(3-4(1-{\cos }^2\theta \left)\right)(2{\cos }^2\theta -1)+(8{\cos }^4\theta -6{\cos }^2\theta )$

$⟹\frac{\sin 5\theta }{\sin \theta }=(-1+4{\cos }^2\theta )(2{\cos }^2\theta -1)+(8{\cos }^4\theta -6{\cos }^2\theta )$

$⟹\frac{\sin 5\theta }{\sin \theta }=-2{\cos }^2\theta +8{\cos }^4\theta +1-4{\cos }^2\theta +8{\cos }^4\theta -6{\cos }^2\theta$

$⟹\frac{\sin 5\theta }{\sin \theta }=16{\cos }^4\theta -12{\cos }^2\theta +1$