Question

$\displaystyle{Prove \ that \ :}\\\\\\\displaystyle{\cos A \ \cos 2A \ \cos 2^2A \ \cos 2^3A \ .. \ .. \ .. \ \cos2^{n-1}A=\dfrac{\sin2^nA}{2^n \ \sin A}}$

Answer 1

Answer:

formula used;

2sinA•cosA=sin2A

please refer to the attachment

thank you

Answer 2

SOLUTION:-

To prove:

cosA.cos2A.cos2²A.cos2³A.cos2^n-1A=sin2^nA/2^nsinA

Proof:

Take L.H.S:

$cosA.cos2A.cos {2}^{2}A.cos {2}^{3} A.cos2 {}^{n - 1} A \\ \frac{1}{2sinA} ([2sinA.cosA]cos2A.cos { {2}^{2} }A.cos {2}^{3} A...cos { {2}^{n - 1} A}) \\ \\ \frac{1}{2sinA} (sin2A.cos2A.cos { {2}^{2} }A.cos {2}^{3} A....cos2 {}^{n - 1} A) \\ \\ \frac{1}{ {2}^{2}sinA }([2sin {2}^{2} A.cos2A].cos {2}^{2} A.cos {2}^{3} A...cos2 {}^{n - 1} A) \\ \\ \frac{1}{ {2}^{2}sinA } (sin {2}^{2} A.cos {2}^{2} A.cos {2}^{3} A...cos {}^{n - 1} A) \\ \\ \frac{1}{ {2}^{3}sinA} (2sin {2}^{2} A.cos {2}^{2} A...cos2 {}^{n - 1} A) \\ \\ \frac{1}{ {2}^{3} sinA} (sin {2}^{3} A.cos {2}^{3} A....cos2 {}^{n - 1} A) \\ . \\ . \\ . \\ . \\ \frac{1}{2 {}^{n - 1}sinA} (sin2 {}^{n - 1} A.cos2 {}^{n - 1} A) \\ \\ \frac{1}{ {2}^{n} sinA} (2sin2 {}^{n - 1} A.cos2 {}^{n - 1} A) \\ \\ \frac{1}{ {2}^{n} sinA} = (sin[2.2 {}^{n - 1} A]) \\ \\ \frac{1}{ {2}^{n}sinA } (sin {2}^{n} A) = R.H.S$

Hence, proved

Hope it helps ☺️