Proving

[hedlund]

Prove that if y = \sin^n{x} then the only value for which y' = \sin{nx} is for n=2. I'm think maybe Euler's formula may be useful ... \sin{x} = \frac{e^{ix}-e^{-ix}}{2i} but I really got no good idea on how to solve it.

[Muzza]

If y = sin^(n)(x), then y' = nsin(x)^(n - 1)cos(x). Suppose n is odd and that y' = sin(nx). Then sin(nx) = nsin(x)^(n - 1)cos(x) even for x = pi/2. The equation reduces into

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.