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2015-05-08T22:27:22+05:30

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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
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