I see that cos(x/2^1) = cos(2*x/2^2) = 1 - 2sin^2(x/2^2).
When k = 1, the product is cos(x/2^1).
When k = 2, the product is (1 - 2sin^2(x/2^2))*cos(x/2^2).
When k = 3, the product is (1 - 4sin^2(x/2^3)cos^2(x/2^3))*(1 - sin^2(x/2^3)*cos(x/2^3).
Does this lead to finding any useful pattern?
RE: Can you solve this integral from MIT's Integration Bee?
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Can you solve this integral from MIT's Integration Bee?
The key step is using
sin(2x) = 2.sinx.cosx
so that
cos(x) = sin(2x)/2sin(x)
and by extension
cos(nx) = sin(2nx)/2sin(nx)
when plugged into the integral product, all the sines cancel out apart from the first and last terms.
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