If y = sin2 x cos2 x , then yn =

Question

If y = sin2 x cos2 x , then yn =  (A) (4n/8) . cos (4x + n π/2) (B)  (-1)n 4n /8) . cos (4x + n π /2)  (C) ( - 4n/8). cos (4x + n π /2) (D) None of these

Tardigrade Admin · Accepted Answer

y = sin2 x cos2x = (1/4) ( sin2 2x) = (1/4) . (1- cos4x/2) = (1/8) (1 - cos4x) = (1/8) - (1/8) cos4x ∴ yn = (1/8) . 4n cos(4x + (nπ/2)) = - (1/8) . 4n cos(4x + (nπ/2))

Q. If $y = sin^{2} x cos^{2} x$ , then $y_{n} =$

$(4^{n} /8) . cos (4 x + n π /2)$

${(- 1)^{n} 4^{n} /8) . cos (4 x + n π /2)}$

$(- 4^{n} /8) . cos (4 x + n π /2)$

None of these

$y = sin^{2} x cos^{2} x = \frac{1}{4} (sin^{2} 2 x)$
$= \frac{1}{4} . \frac{1 - c o s 4 x}{2} = \frac{1}{8} (1 - cos 4 x)$
$= \frac{1}{8} - \frac{1}{8} cos 4 x ∴ y_{n} = \frac{1}{8} . 4^{n} cos (4 x + \frac{nπ}{2})$
$= - \frac{1}{8} . 4^{n} cos (4 x + \frac{nπ}{2})$

Q. If y=sin2xcos2x , then yn​=

(4n/8).cos(4x+nπ/2)

{(−1)n4n/8).cos(4x+nπ/2)}

(−4n/8).cos(4x+nπ/2)