MHT CET · Maths · Trigonometric Ratios & Identities
The value of \(\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)\) is
- A \(\frac{1}{8}\)
- B \(\frac{-1}{8}\)
- C \(\frac{1}{16}\)
- D \(\frac{-1}{16}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{8}\)
Step-by-step Solution
Detailed explanation
\(\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\)
\(=\left(1+\cos \frac{\pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right)\) \(\ldots[\because \cos (\pi-\theta)=-\cos \theta]\)
\(=\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right) \)
\( =\sin ^2 \frac{\pi}{8} \sin ^2 \frac{3 \pi}{8} \)
\( =\frac{1}{4}\left(2 \sin \frac{\pi}{8} \cdot \sin \frac{3 \pi}{8}\right)^2 \)
\( =\frac{1}{4}\left(\cos \frac{\pi}{4}-\cos \frac{\pi}{2}\right)^2 \)
\( =\frac{1}{8}\)
\(=\left(1+\cos \frac{\pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right)\) \(\ldots[\because \cos (\pi-\theta)=-\cos \theta]\)
\(=\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right) \)
\( =\sin ^2 \frac{\pi}{8} \sin ^2 \frac{3 \pi}{8} \)
\( =\frac{1}{4}\left(2 \sin \frac{\pi}{8} \cdot \sin \frac{3 \pi}{8}\right)^2 \)
\( =\frac{1}{4}\left(\cos \frac{\pi}{4}-\cos \frac{\pi}{2}\right)^2 \)
\( =\frac{1}{8}\)
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