<s>[INST] Large powers of sine appear Gaussian -- why? As part of approximating an integral, I have noticed that $\sin^k(x), x\in[0, \pi]$ look almost identical to $\exp\left(-\frac{k}{2}(x-\frac{\pi}{2})^2\right)$ once $k$ is large enough (in practice, the two equations are visually identical for $k\geq 15$). As an example, see this picture of the two functions for $k=10$:

Can anybody explain why these functions practically are identical?
Edit: I should mention that the expression $\exp\left(-\frac{k}{2}(x-\frac{\pi}{2})^2\right)$ is derived by a 2nd order Taylor expansion of $\log\sin^k(x)$ around the mode $x_0 = \frac{\pi}{2}$, i.e. a Laplace approximation.
 [/INST] This is too long for a comment.
Semiclassical made a very good comment.
Let us use Taylor series around $x=\frac \pi 2$
$$\sin(x)=1-\frac{1}{2} \left(x-\frac{\pi }{2}\right)^2+\frac{1}{24} \left(x-\frac{\pi
   }{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^6\right)$$ $$\sin^k(x)=1-\frac{1}{2} k \left(x-\frac{\pi }{2}\right)^2+\frac{1}{24} k (3 k-2)
   \left(x-\frac{\pi }{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^6\right)$$ On the other hand $$e^{-\frac{1}{2} k \left(x-\frac{\pi }{2}\right)^2}=1-\frac{1}{2} k \left(x-\frac{\pi }{2}\right)^2+\frac{1}{8} k^2 \left(x-\frac{\pi
   }{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^6\right)$$ Computing the difference $$\sin^k(x)-e^{-\frac{1}{2} k \left(x-\frac{\pi }{2}\right)^2}=-\frac{1}{12} k \left(x-\frac{\pi }{2}\right)^4+O\left(\left(x-\frac{\pi
   }{2}\right)^6\right)$$ AT least, they are very close in the area around the maximum.
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