Trigonometrical Ratios And Identities Ques 17
$(1+\cos \frac{\pi}{8} ) \quad (1+\cos \frac{3 \pi}{8}) \quad (1+\cos \frac{5 \pi}{8}) \quad (1+\cos \frac{7 \pi}{8})$ is equal to
(a) $\frac{1}{2}$
(b) $\cos \frac{\pi}{8}$
(c) $\frac{1}{8}$
(d) $\frac{1+\sqrt{2}}{2 \sqrt{2}}$
$(1984,3 M)$
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Answer:
Correct Answer: 17.(c)
Solution:
Formula:
- Given expression $=$
$ \begin{aligned} (1 & +\cos \frac{\pi}{8}) \quad (1+\cos \frac{3 \pi}{8} )\quad (1+\cos \frac{5 \pi}{8} )\quad ( 1+\cos \frac{7 \pi}{8}) \\ & =(1+\cos \frac{\pi}{8}) \quad (1+\cos \frac{3 \pi}{8} )\quad (1-\cos \frac{3 \pi}{8} )\quad (1-\cos \frac{\pi}{8}) \end{aligned} $
$ \begin{aligned} & =(1-\cos ^{2} \frac{\pi}{8} )\quad (1-\cos ^{2} \frac{3 \pi}{8}) \\ & =\frac{1}{4} (2-1-\cos \frac{\pi}{4}) \quad (2-1-\cos 3 \frac{\pi}{4}) \\ & =\frac{1}{4} (1-\cos \frac{\pi}{4}) \quad (1-\cos 3 \frac{\pi}{4}) \\ & =\frac{1}{4} (1-\frac{1}{\sqrt{2}}) \quad (1+\frac{1}{\sqrt{2}})=\frac{1}{4} (1-\frac{1}{2})=\frac{1}{8} \end{aligned} $
BETA
