# Conclusion from trigonometric identity

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.

• $\sin x=\sin(\pi-x)$, and $x^{2k}=(-x)^{2k}$. – Lucian Feb 15 '14 at 20:32
• @Lucian I know that, but how that would help me? – Stefan4024 Feb 15 '14 at 20:38

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 $\beta=1.3608$, $g(\cos(\alpha))\approx 0.1369 \approx g(\cos(\beta))$.

• nice answer @Chuck – Semsem Feb 15 '14 at 20:47
• In the first step, why $\sin\alpha$ and $\sin\beta$? Do you mean $\cos\alpha$ and $\cos\beta$? – user21467 Feb 15 '14 at 20:50
• About the example $\alpha=\frac\pi4$ and $\beta=\frac{3\pi}4$: these are angles in a (nondegenerate?) triangle, so we know $\alpha+\beta\in(0,\pi)$. – user21467 Feb 15 '14 at 20:53
• @ChuckJia As far as I know $\sin \alpha$ and $\sin \beta$ should be replaced by their respective cosine function, right? That completely changes the complexity of the problem. – Stefan4024 Feb 15 '14 at 20:58
• @StevenTaschuk You are right. But I think your original statement may not hold. I'm editing my answer to show that. – Capricornus Feb 15 '14 at 21:45

The terms of the equation are undefined if the denominators are zero, so we exclude that eventuality.
Take the square root of both sides, and make substitution $\alpha/2=a$, $\beta/2=b$, apply double-angle formula on the left to get $$\frac{\sin^{50}a\cos^{50}a}{\sin^{50}b\cos^{50}b}=\frac{\sin a}{\sin b}\,,$$ after canceling the $2^{50}$ from top and bottom on the left. Clearing of fractions, we get $$\sin^{50}a\cos^{50}a\sin b=\sin^{450}b\cos^{50}b\sin a\,,$$ and under the assumption that $\sin a\sin b\ne0$, we get $\sin^{49}a\cos^{50}a=\sin^{49}b\cos^{50}b$. In other words, we’re asking about the function $f(x)=\sin^{49}x\cos^{50}x$, and in particular whether there are “interesting” pairs of values $(x,x')$ where $f(x)=f(x')$. For instance, are there $x,x'$ with $0<x<x'<\pi/2$? Certainly there are, since $f$ is positive on the open interval $\langle0,\pi/2\rangle$, zero at the endpoints, and continuous, so if $\eta$ is a positive value that’s less than the maximum of $f$ on $[0,\pi/2]$, there are two $x$ values such that $f(x)=\eta$.