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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

  1. A12\frac{1}{2}
  2. B12+122\frac{1}{2}+\frac{1}{2 \sqrt{2}}
  3. C12−122\frac{1}{2}-\frac{1}{2 \sqrt{2}}
  4. D18\frac{1}{8}

Correct answer

D. 18\frac{1}{8}

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⁡7π8=cos⁡[π−π8]=−cos⁡π8 and  cos⁡5π8=cos⁡[π−3π8]=−cos⁡3π8∴=[1+cos⁡π8][1+cos⁡3π8][1−cos⁡π8][1−cos⁡3π8]=[1−cos⁡2π8][1−cos⁡23π8]=sin⁡2π8⋅sin⁡23π8=14[2sin⁡2π8⋅2sin⁡23π8]=14[(1−cos⁡π4)(1−cos⁡3π4)]=14[\lleft(1−12)\lleft(1+12)]=18\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}$$
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