Question

If $\sin 6\theta = 32\cos^{5}\theta.\sin\theta - 32\cos^{3}\theta\sin\theta + 3x,$ then x =

• A. $\cos\theta$
• B. $\cos 2\theta$
• C. $\sin\theta$
• D. $\sin 2\theta$

Answer 1

Answer: (D)

$\sin 2\theta$

Answer 2

$\sin 6\theta = 2\sin 3\theta.\cos 3\theta$ =

$2\lbrack 3\sin\theta - 4\sin^{3}\theta\rbrack\lbrack 4\cos^{3}\theta - 3\cos\theta\rbrack$ =$\tan A = \frac{3}{2}\tan B = \frac{3}{2}t$ = $32\cos^{5}\theta.\sin\theta - 32\cos^{3}\theta.\sin\theta + 3\sin 2\theta$On comparing,

$2\tan A = 3\tan B$

Trick : Put $\theta = 0^{o}$, then $x = 0$. So, option (3) and (4) are correct.

Now put $\theta = 30^{o}$, then$x = \frac{\sqrt{3}}{2}$. Therefore, Only option (4) is correct.