sin^4 (π)/8+sin^4 (2 π)/8+sin^4 (3 π)/8+sin^4 (4 π)/8 +sin^4 (5 π)/8+sin^4 (6 π)/8+sin^4 (7 π)/8=

Question

MARKS Content Team · Accepted Answer

Correct answer: (C) 3. sin^4 (π)/8+sin^4 (2 π)/8+ sin^4 (3 π)/8+sin^4 (4 π)/8 +sin^4 (5 π)/8+sin^4 (6 π)/8+sin^4 (7 π)/8 = [sin^4 (π)/8+sin^4 (3 π)/8+sin^4 (5 π)/8+sin^4 (7 π)/8 ] + [sin^4 (2 π)/8+sin^4 (4 π)/8+sin^4 (6 π)/8 ] = [sin^4 (π)/8+sin^4 (3 π)/8+cos^4 ((π)/2-(5 π)/8 ) . .+cos^4 ((π)/2-(7 π)/8 ) ]+ [sin^4 (π)/4+sin^4 (π)/2+sin^4 (3 π)/4 ] = [sin^4 (π)/8+sin^4 (3 π)/8+cos^4 (-(π)/8 )+cos^4 ((-3 π)/8 ) ] + [sin^4 (π)/4+sin^4 (π)/2+sin^4 ((π)/2+(π)/4 ) ] = [sin^4 (π)/8+sin^4 (3 π)/8+cos^4 (π)/8+cos^4 (3 π)/8 ] + [sin^4 (π)/4+sin^4 (π)/2+cos^4 (π)/4 ] = [ (sin^4 (π)/8+cos^4 (π)/8 )+ (sin^4 (3 π)/8+cos^4 (3 π)/8 ) ] + [ (1/(√2) )^4+(1)^4+ (1/(√2) )^4 ] = [ (1-2 sin^2 (π)/8 cos^2 (π)/8 )+ (1-2 sin^2 (3 π)/8 cos^2 (3 π)/8 ) ] + [1/4+1+1/4 ] = [2-1/2 4 sin^2 (π)/8 cos^2 (π)/8-1/2 4 sin^2 (3 π)/8 cos^2 (3 π)/8 ]+ [1/2+1 ] = [2-1/2 (sin^2 (π)/4 )-1/2 (sin^2 (3 π)/4 ) ]+3/2 =2-1/2 (1/2 )-1/2 (1/2 )+3/2=2-1/4-1/4+3/2=3

$\begin{aligned} & \sin ^4 \frac{\pi}{8}+\sin ^4 \frac{2 \pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{4 \pi}{8} \\ & +\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{6 \pi}{8}+\sin ^4 \frac{7 \pi}{8}=\end{aligned}$

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