Trigonometrical Ratios And Identities Ques 36

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: 36.(c)

Solution:

Formula:

Double Angle Identities:

  1. 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 $\left.n=9\right]$

$=\frac{1}{2^{9}} \sin [\frac{\pi}{2}]=\frac{1}{2^{9}}=\frac{1}{512}$



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