# If $a+b+c=\pi$ and $\cot t=\cot a+\cot b+\cot c$, show $\sin^3t=\sin(a-t)\sin(b-t)\sin(c-t)$

I have a problem from a mathematics book:

If $$\alpha + \beta +\gamma = \pi \tag{1}$$ and $$\cot \theta = \cot\alpha + \cot \beta + \cot \gamma, 0 < \theta < \frac{\pi}{2}\tag{2}$$ show that $$\sin^{3}\theta = \sin(\alpha - \theta)\sin(\beta - \theta)\sin(\gamma - \theta)\tag{3}$$

The only way I can think of doing this is by using brute force (multiplying out all the terms on the right of equation (3) ). There must be some solution to this that is not as complicated. I have tried using the identity that if $$\alpha + \beta +\gamma = \pi$$ then $$\tan\alpha+\tan\beta+\tan\gamma=\tan\alpha\tan\beta\tan\gamma$$, but it does not fit as (2) uses cotangents instead of tangents.

Somehow, the $$\alpha$$, $$\beta$$ and $$\gamma$$ all disappear in (3). My question is: how can I solve this, and is there any way other than brute force to solve this?

This is the Brocard angle of a triangle, conventionally denoted by $$\omega$$ rather than $$\theta$$.
Anyway, by expanding the compound angle formula for sine, you get \begin{align*} \frac{\sin(\alpha-\theta)}{\sin\theta}&=\frac{\sin\alpha\cos\theta-\cos\alpha\sin\theta}{\sin\theta}\\ &=\sin\alpha(\cot\theta-\cot\alpha)\\ &=\sin\alpha(\cot\beta+\cot\gamma)\\ &=\sin\alpha\cdot\frac{\cos\beta\sin\gamma+\sin\beta\cos\gamma}{\sin\beta\sin\gamma}\\ &=\frac{\sin\alpha}{\sin\beta\sin\gamma}\cdot(\cos\beta\sin\gamma+\sin\beta\cos\gamma)\\ &=\frac{\sin\alpha}{\sin\beta\sin\gamma}\cdot\sin(\beta+\gamma)\\ &=\frac{\sin^2\alpha}{\sin\beta\sin\gamma} \end{align*} and similar. So taking their product gives equation (3).