Trigonometrical Ratios and Identities 1 Question 6
6. The value of $\cos \frac{\pi}{2^{2}} \cdot \cos \frac{\pi}{2^{3}} \ldots \ldots \cos \frac{\pi}{2^{10}} \cdot \sin \frac{\pi}{2^{10}}$ is
(a) $\frac{1}{1024}$
(b) $\frac{1}{2}$
(c) $\frac{1}{512}$
(d) $\frac{1}{256}$
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Answer:
Correct Answer: 6. (a, b, c, d)
Solution:
- We know that, $\cos \alpha \cdot \cos (2 \alpha) \cos \left(2^{2} \alpha\right) \ldots \cos \left(2^{n-1} \alpha\right)=\frac{\sin \left(2^{n} \alpha\right)}{2^{n} \sin \alpha}$
$\therefore \cos \frac{\pi}{2^{2}} \cdot \cos \frac{\pi}{2^{3}} \ldots \cos \frac{\pi}{2^{10}} \cdot \sin \frac{\pi}{2^{10}}$
$=\frac{\sin \frac{\pi}{2^{10}} 2^{9}}{2^{9} \sin \frac{\pi}{2^{10}}} \sin \frac{\pi}{2^{10}} \quad\left[\because\right.$ here,$\alpha=\frac{\pi}{2^{10}}$ and $\left.n=9\right]$
$=\frac{1}{2^{9}} \sin \frac{\pi}{2}=\frac{1}{2^{9}}=\frac{1}{512}$
