Expand sine^2 around a large number

Question

Hi how do I expand $(\sin(x))^2$ for $x\gg1$? This maybe a dumb question but im wondering whether there exists such a method. Thanks

Answer 1

Sure. The simplest approach is to use the fact that $\sin(x + 2\pi n) = \sin(x)$ for any integer $n$. So, for example, since $\sin(x) \approx x$ for small $x$, $\sin^2(48\pi + 0.001) \approx 0.001^2$.

So, in other words, take your expansion for $\sin^2(x)$ around $0$, and then replace $x$ with $x - 2\pi n$ for some choice of $n$; you'll have a good expansion of $\sin^2(x)$ around $2\pi n$.

Answer 2

You can always use the fact that $\sin^2x=\frac {1-\cos 2x}2$. If $x$ is near an odd multiple of $\pi$ this becomes $\sin^2((2k+1)\pi+x)=\frac {1-\cos 2x}2$ and use the Taylor series for $\cos 2x$.