Question

If $y = \sin^2 x \cos^2 x$ , then $y_n = $

• A. $(4^n/8) .\cos (4x + n\, \pi/2)$
• B. $\{ (-1)^n 4^n /8) . \cos (4x + n\, \pi /2)\}$
• C. $( - 4^n/8). \cos (4x + n \,\pi /2)$
• D. None of these

Answer 1

Answer: (C)

$y =\sin^{2} x \cos^{2}x = \frac{1}{4} \left(\sin^{2} 2x\right)$
$ = \frac{1}{4} . \frac{1-\cos4x}{2} = \frac{1}{8} \left(1 - \cos4x\right) $
$= \frac{1}{8} - \frac{1}{8} \cos4x \therefore y_{n} = \frac{1}{8} . 4^{n} \cos\left(4x + \frac{n\pi}{2}\right)$
$= - \frac{1}{8} . 4^{n} \cos\left(4x + \frac{n\pi}{2}\right)$