Trigonometrical Ratios and Identities 1 Question 15
15. $1+\cos \frac{\pi}{8} \quad 1+\cos \frac{3 \pi}{8} \quad 1+\cos \frac{5 \pi}{8} \quad 1+\cos \frac{7 \pi}{8}$ 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: 15. (b)
Solution:
- Given expression $=$
$$ \begin{aligned} 1 & +\cos \frac{\pi}{8} \quad 1+\cos \frac{3 \pi}{8} \quad 1+\cos \frac{5 \pi}{8} \quad 1+\cos \frac{7 \pi}{8} \\ & =1+\cos \frac{\pi}{8} \quad 1+\cos \frac{3 \pi}{8} \quad 1-\cos \frac{3 \pi}{8} \quad 1-\cos \frac{\pi}{8} \end{aligned} $$
$$ \begin{aligned} & =1-\cos ^{2} \frac{\pi}{8} \quad 1-\cos ^{2} \frac{3 \pi}{8} \\ & =\frac{1}{4} 2-1-\cos \frac{\pi}{4} \quad 2-1-\cos 3 \frac{\pi}{4} \\ & =\frac{1}{4} 1-\cos \frac{\pi}{4} \quad 1-\cos 3 \frac{\pi}{4} \\ & =\frac{1}{4} 1-\frac{1}{\sqrt{2}} \quad 1+\frac{1}{\sqrt{2}}=\frac{1}{4} 1-\frac{1}{2}=\frac{1}{8} \end{aligned} $$
