Let $\alpha$ and $\beta$ be angles in triangle, i.e $\alpha, \beta \in \left(0,\pi\right)$ can we conclude that $\alpha = \beta$ if the following statement is true:
$$\left(\frac{\sin \alpha}{\sin \beta}\right)^{100}= \frac{\sin^2 \frac {\alpha}{2}}{\sin^2 \frac{\beta}{2}}$$
For me this looks too obvious, but I'm unable to solve it for hours.
We know that $$ \left(\frac{\sin\alpha}{\sin\beta}\right)^{100}=\frac{1-\cos\alpha}{1-\cos\beta} $$ Then $$ \frac{(\sin\alpha)^{100}}{1-\cos\alpha}=\frac{(\sin\beta)^{100}}{1-\cos\beta} $$ Let $$ f(x):=\frac{(\sin x)^{100}}{1-\cos x} $$ Then $$ f'(x)=\frac{(\sin x)^{99}}{(1-\cos x)^2}\left( 100\cos x - 99 \cos^2 x -1 \right) $$ Let $g(x):=100y - 99 y^2 -1$.
When $\cos (x)<0$, $$ 100\cos x - 99 \cos^2 x -1<0 $$
When $\cos x>0$, $$ g(\frac{50}{99})=\frac{2401}{99}\approx 24.25 $$ In fact when $x\in \arccos(1),\arccos(\frac{50}{99})$, $\cos x\in (\frac{50}{99},1)$, and thus $g(\cos(x))$ decreases. When $x\in (\arccos(\frac{50}{99}),\arccos(-1)$, $g(\cos(x))$ increases. So we know that $g(\cos(x))$ is not injective. Thus $\alpha$ may not equal $\beta$.
Numerically, we can find examples. Let $\alpha=1.7608$, and $\beta=1.3608$, $g(\cos(\alpha))\approx 0.1369 \approx g(\cos(\beta))$.