I'm supposed to prove by induction the equality n∏k=1cosx2k=sinx2nsinx2n I've shown the base case for n=1, assumed it is valid for n then tried extending for n+1. I've used trigonometric identities to obtain the expression (2−2cosx2n)cos2x2k=sin2x2k and it seems to me that I'm close, but I'm stuck here. Any help please?
Answer
To get the proof consider the identity cos(x)sin(x)=12sin(2x).
Now the induction step leads to
n+1∏k=1cos(x/2k)=cos(x/2n+1)n∏k=1cos(x/2k).
On the last term you use the induction hypothesis.
Using the proposed equality the first term is equal to
cos(x/2n+1)=sin(x/2n)2sin(x/2n+1)
Simplifying should yield the result.
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