Question

Class 11th
Prove that
cosA.cos2A.cos2²A-----------------Cos$2^{n+1} A$ = $\frac{sin2^{n}A }{ 2^{n}sinA }$

Answer 1

Induction works nicely for this proof

Base case :
$\cos2^0A=\frac{2\sin{A}\cos{A}}{2\sin{A}}=\frac{\sin2^1A}{2^1\sin{A}}~~\checkmark$

Induction hypothesis :
Let $k\in\mathbb{N}$ be an integer greater than $0$ and assume that the proposition is true for $n=k$. That is
$\cos{A}\cdot\cos2A\cdot\cos2^2A\cdots\cos2^{k-1}A=\frac{\sin2^kA}{2^k\sin{A}}$

Induction step :
$\cos{A}\cdot\cos2A\cdot\cos2^2A\cdots\cos2^{k-1}A\cdot\cos2^{k}A=\frac{\sin2^kA}{2^k\sin{A}}\cdot\cos2^{k}A$

$=\frac{2\sin2^kA\cos2^kA}{2\cdot2^k\sin{A}}$

$=\frac{\sin2^{k+1}A}{2^{k+1}\sin{A}}\\\blacksquare$

Answer 2

Answer:

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