Question

If the value of
$cos\frac{\pi }{2^2}.cos\frac{\pi }{2^3}\cdot \cdots \cdot cos\frac{\pi }{2^{10}}\cdot sin\frac{\pi }{2^{10}}=\frac{1}{2^k}$, then $k$ is equal to

Answer 1

Answer: 9

$A=cos\frac{\pi }{2^2}\cdot cos\frac{\pi }{2^3}\dots cos\frac{\pi }{2^{10}}\cdot sin\frac{\pi }{10}$
$=\frac{1}{2}\left(cos\frac{\pi }{2^2}\cdot cos\frac{\pi }{2^3}\dots cos\frac{\pi }{2^9}sin\frac{\pi }{2^9}\right)$
$=\frac{1}{2^8}\left(cos\frac{\pi }{2^2}\cdot sin\frac{\pi }{2^2}\right)=\frac{1}{2^9}sin\frac{\pi }{2}$
$=\frac{1}{512}=\frac{1}{2^9}$