# How to proof this trig equality???(question 1 : solved)(question 2 : solved)

question 1:

We know : $$\left\{ \begin{array}{ll} \alpha \ + \beta\ +\theta\ = \pi\\ \alpha\ = \frac{\pi}{2} \\ \end{array} \right.$$

How to prove this :

$$\sin(\alpha)\sin(\beta)\sin(\alpha\ -\ \beta) + \sin(\beta)\sin(\theta)\sin(\beta\ -\ \theta) + \sin(\theta)\sin(\alpha)\sin(\theta\ -\ \alpha)+ \\ \sin(\alpha\ -\ \beta)\sin(\beta\ -\ \theta)\sin(\theta\ -\ \alpha) = 0$$

question 2:

And if we know that : $$\left\{ \begin{array}{ll} \alpha \ + \beta\ +\theta\ = \pi\\ \cot(\alpha) \ + \cot(\beta)\ + \cot(\theta) \ = 2 \\ \end{array} \right.$$

How to prove this:

$$1+\cos(\alpha)\cos(\beta)\cos(\theta) = 2\times \sin(\alpha)\sin(\beta)\sin(\theta)$$

Thanks a lot.

• Replace $\theta$ with $\dfrac\pi2-\beta$ – lab bhattacharjee Mar 4 '16 at 6:04
• @K.K.McDonald $sin{\alpha}=1$ – user41736 Mar 4 '16 at 6:14
• Wait. In the first one do we know both of those statements are true??? If alpha = pi/2 our work is 5/6 done! – fleablood Mar 4 '16 at 6:54
• OK dude for answering this question u can use a,b,c instead of α,β,θ. – K.K.McDonald Mar 4 '16 at 6:55
• yeah in first question both of these statements are true. – K.K.McDonald Mar 4 '16 at 6:56

The relation $\alpha + \beta + \theta = \pi$ implies that $\cos (\alpha + \beta + \theta) = -1$. But \begin{align} \cos(\alpha + \beta + \theta) &= \cos\alpha \cos(\beta + \theta) - \sin \alpha\sin(\beta + \theta) \\ &= \cos(\alpha)(\cos\beta\cos\theta - \sin\beta \sin\theta) - \sin\alpha(\sin\beta \cos\theta + \sin\theta \cos\beta) \\ &= \cos\alpha \cos\beta \cos\theta - \cos\alpha\sin\beta\sin\theta - \sin\alpha\sin\beta\cos\theta - \sin\alpha\sin\theta\cos\beta, \end{align} so $$1 +\cos\alpha\cos\beta\cos\theta = \cos\alpha\sin\beta\sin\theta + \sin\alpha\sin\beta\cos\theta + \sin\alpha\sin\theta\cos\beta.$$ Therefore $$\frac{1}{\sin\alpha\sin\beta\sin\theta} + \cot\alpha\cot\beta\cot\theta = \cot\alpha + \cot\beta + \cot\theta = 2$$ and the result follows.