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- Thread starter hedlund
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sin(n * pi/2) = nsin(pi/2)^(n - 1)cos(pi/2)

<=>

sin(n * pi/2) = 0.

But n = 2k + 1 for some integer k, so sin(n * pi/2) = sin( (2k + 1)pi/2 ) = sin(kpi + pi/2) = cos(k * pi). But that is never equal to zero. Contradiction.

I don't know how to handle the case when n is even and > 2.

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