Prove that $n\sin\frac{2\pi}{n}-\frac{1}{4}n\sin\frac{4\pi}{n}>\pi$ (corrected inequation)

Question

Prove that Prove that $n\sin\frac{2\pi}{n}-\frac{1}{4}n\sin\frac{4\pi}{n}>\pi$ algebraically or geometrically.

$n\sin\frac{2\pi}{n}-n\sin\frac{\pi}{n}$ means the area of a regular n-gon + the area difference between a regular n-gon and a regular n/2-gon.

Thanks.

Update: Sorry I gave the wrong inequation at first time; I corrected it.

Answer 1

u mean < instead of >?

$n \sin \frac{2 \pi}{n} - n \sin \frac{\pi}{n} = 2 n \sin\frac{\pi}{2n} \cos \frac{3 \pi}{2n} < 2n * \frac{\pi}{2n}* 1 = \pi$

Answer 2

As Kanye West, I wonder if you inequality is properly written.

Suppose we define $x=\frac{\pi}{n}$. So we need to compare $\Big(\sin(2x)-\sin(x)\Big)$ to $x$. If $x$ is small (then $n$ is large), Taylor expansion gives $$\sin(2x)-\sin(x)=x-\frac{7 x^3}{6}+\frac{31 x^5}{120}+O\left(x^7\right)$$ which is smaller than $x$.

In the same manner, if you plot $$f(n)=n\sin\frac{2\pi}{n}-n\sin\frac{\pi}{n}-\pi$$ as a function of $n$, you will see than $f(n) \lt 0$ for any value of $n$ and that $0^-$ is the asymptote.

Answer 3

Divide by $n$, and let $\displaystyle x={2\pi\over n}$; then we want to show $$0<x\le{\pi\over2}{\rm\quad implies\quad }\sin x-{1\over4}\sin2x>{x\over2}$$ So let $$f(x)=\sin x-{1\over4}\sin2x-{x\over2}$$ We have $f(0)=0$, and, for $\displaystyle0<x<{\pi\over2}$ $$f'(x)=\cos x-{1\over2}\cos2x-{1\over2}=\cos x-\cos^2x=\cos x(1-\cos x)>0$$ So $f(x)>0$ for $\displaystyle0<x\le{\pi\over2}$, and we're done.