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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2022 · Mathematics · Trigonometric Ratios & Identities

What is the value of ^4 (7 π)/(8)+ ^4 (5 π)/(8) ?

What is the value of cos⁡47π8+cos⁡45π8\cos ^4 \frac{7 \pi}{8}+\cos ^4 \frac{5 \pi}{8} ?

Options

  1. A32\frac{3}{2}
  2. B34\frac{3}{4}
  3. C38\frac{3}{8}
  4. D316\frac{3}{16}

Correct answer

B. 34\frac{3}{4}

Step-by-step solution

cos⁡47π8+cos⁡45π8=(cos⁡27π8)2+(cos⁡25π8)2=(cos⁡27π8−cos⁡25π8)2+2cos⁡27π8⋅cos⁡25π8=[(−1)sin⁡(7π8+5π8)sin⁡(7π8−5π8)]2+12[2cos⁡7π8⋅cos⁡5π8]2=[sin⁡3π2⋅sin⁡π4]2+12[cos⁡3π2+cos⁡π4]2\begin{aligned} & \cos ^4 \frac{7 \pi}{8}+\cos ^4 \frac{5 \pi}{8}=\left(\cos ^2 \frac{7 \pi}{8}\right)^2+\left(\cos ^2 \frac{5 \pi}{8}\right)^2 \\ = & \left(\cos ^2 \frac{7 \pi}{8}-\cos ^2 \frac{5 \pi}{8}\right)^2+2 \cos ^2 \frac{7 \pi}{8} \cdot \cos ^2 \frac{5 \pi}{8} \\ = & {\left[(-1) \sin \left(\frac{7 \pi}{8}+\frac{5 \pi}{8}\right) \sin \left(\frac{7 \pi}{8}-\frac{5 \pi}{8}\right)\right]^2+\frac{1}{2}\left[2 \cos \frac{7 \pi}{8} \cdot \cos \frac{5 \pi}{8}\right]^2 } \\ = & {\left[\sin \frac{3 \pi}{2} \cdot \sin \frac{\pi}{4}\right]^2+\frac{1}{2}\left[\cos \frac{3 \pi}{2}+\cos \frac{\pi}{4}\right]^2 }\end{aligned}=\left[(-1) \cdot \frac{1}{\sqrt{2}}\right]^2+\frac{1}{2}\left[0+\frac{1}{\sqrt{2}}\right]^2=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$
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