![this is a very dicy problem. It would be great to go into details of how to prove it using induction or any other alternate way is highly appreciated.][1]
$$\sin(x)\cos(x)\cos(2x)\cos(4x)\cos(8x)...\cos(2^nx)=\frac{\sin(2^{n+1}x)}{2^{n+1}}$$
this is a very dicy problem. It would be great to go into details of how to prove it using induction or any other alternate way is highly appreciated.
Answer
Try : $\sin(2a) = 2*\sin(a)*\cos(a)$
By induction: the latter proves the case n=0
Let that be true for n:
$sin(x)*cos(x)*...*cos(2^nx) = \frac{sin(2^{n+1}x)}{2^{n+1}}$
=> $sin(x)*cos(x)*...*cos(2^nx)*cos(2^{n+1}x) = \frac{sin(2^{n+1}x)}{2^{n+1}}*cos(2^{n+1}x) = \frac{1}{2^{n+1}}*\frac{sin(2^{n+2}x)}{2} $
Here I used the relation I gave above with $a= 2^{n+1}x$
=> $sin(x)*cos(x)*...*cos(2^nx)*cos(2^{n+1}x) = \frac{sin(2^{n+2}x)}{2^{n+2}} $
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