NDA 2016
Exam NDA
Year 2016
Subject Mathematics
Chapter Trigonometric Ratios & Identities
NDA 2016 · Mathematics · Trigonometric Ratios & Identities
What is the value of (1+ (π)/(8))(1+ (3 π)/(8))(1+ (5 π)/(8))(1+ (7...
What is the value of
\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right) ?
Options A 1 2 \frac{1}{2} 2 1 B 1 2 + 1 2 2 \frac{1}{2}+\frac{1}{2 \sqrt{2}} 2 1 + 2 2 1 C 1 2 − 1 2 2 \frac{1}{2}-\frac{1}{2 \sqrt{2}} 2 1 − 2 2 1 D 1 8 \frac{1}{8} 8 1
Correct answer D. 1 8 \frac{1}{8} 8 1
Step-by-step solution
\left[1+\cos \frac{\pi}{8}\right] \left[ 1+\cos
\frac{3 \pi}{8}\right]\left[1+\cos \frac{5 \pi}{8}\right] \left[ 1+
\frac{\cos 7 \pi}{8}\right]
We have, cos 7 π 8 = cos [ π − π 8 ] = − cos π 8 and cos 5 π 8 = cos [ π − 3 π 8 ] = − cos 3 π 8 ∴ = [ 1 + cos π 8 ] [ 1 + cos 3 π 8 ] [ 1 − cos π 8 ] [ 1 − cos 3 π 8 ] = [ 1 − cos 2 π 8 ] [ 1 − cos 2 3 π 8 ] = sin 2 π 8 ⋅ sin 2 3 π 8 = 1 4 [ 2 sin 2 π 8 ⋅ 2 sin 2 3 π 8 ] = 1 4 [ ( 1 − cos π 4 ) ( 1 − cos 3 π 4 ) ] = 1 4 [ \lleft ( 1 − 1 2 ) \lleft ( 1 + 1 2 ) ] = 1 8 \begin{aligned}& \cos \frac{7 \pi}{8}=\cos \left[\pi-\frac{\pi}{8}\right]=-\cos \frac{\pi}{8}\\
& \text{\text{ and } } \cos \frac{5 \pi}{8}=\cos \left[\pi-\frac{3 \pi}{8}\right]=-\cos \frac{3 \pi}{8}\\
& \therefore=\left[1+\cos \frac{\pi}{8}\right] \left[1+\cos \frac{3 \pi}{8}\right]\left[1-\cos \frac{\pi}{8}\right]\left[1-\cos \frac{3 \pi}{8}\right]\\
& =\left[1-\cos ^2 \frac{\pi}{8}\right]\left[1-\cos ^2 \frac{3 \pi}{8}\right]=\sin ^2 \frac{\pi}{8} \cdot \sin ^2 \frac{3 \pi}{8}\\
& =\frac{1}{4} \left[2 \sin ^2 \frac{\pi}{8} \cdot2 \sin ^2 \frac{3 \pi}{8}\right]\\
& =\frac{1}{4}\left[ \left(1-\cos \frac{\pi}{4}\right)\left(1-\cos \frac{3 \pi}{4}\right)\right]\\
& =\frac{1}{4}\left[\lleft(1-\frac{1}{\sqrt{2}}\right)\lleft(1+\frac{1}{\sqrt{2}})]=\frac{1}{8}
\end{aligned} cos 8 7 π = cos [ π − 8 π ] = − cos 8 π and cos 8 5 π = cos [ π − 8 3 π ] = − cos 8 3 π ∴ = [ 1 + cos 8 π ] [ 1 + cos 8 3 π ] [ 1 − cos 8 π ] [ 1 − cos 8 3 π ] = [ 1 − cos 2 8 π ] [ 1 − cos 2 8 3 π ] = sin 2 8 π ⋅ sin 2 8 3 π = 4 1 [ 2 sin 2 8 π ⋅ 2 sin 2 8 3 π ] = 4 1 [ ( 1 − cos 4 π ) ( 1 − cos 4 3 π ) ] = 4 1 [ \lleft ( 1 − 2 1 ) \lleft ( 1 + 2 1 )] = 8 1 $$
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