Given that $\alpha + \beta - \gamma = \pi$, prove that $\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = 2 \sin \alpha \sin \beta \cos \gamma$. I am told:
$$\alpha + \beta - \gamma = \pi$$
And I have to prove:
$$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = 2 \sin \alpha \sin \beta \cos \gamma$$
What should I be looking for? I kept trying to take the sine of bots sides and use the formulas:
$$\sin(a + b) = \sin a \cos b + \sin b \cos a$$
$$\sin(a-b) = \sin a \cos b - \sin b \cos a$$
but got nowhere. Then I tried using the formulas:
$$\sin a + \sin b = 2 \sin \bigg ( \dfrac{a + b}{2} \bigg ) \cos\bigg ( \dfrac{a - b}{2} \bigg )$$
$$\sin a - \sin b = 2 \cos \bigg ( \dfrac{a + b}{2} \bigg ) \sin \bigg ( \dfrac{a - b}{2} \bigg )$$
But again, I got nowhere. Can you give me a hint? At least what should I be looking for? What should be my strategy? Everything that I did felt just random, while kind of hoping that everything would just magically turn into the desired result. What is the strategy for this kind of problem?
 A: Hint:
Use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $
$$\sin^2\beta-\sin^2\gamma=\sin(\beta+\gamma)\sin(\beta-\gamma)=\sin(\beta+\gamma)\sin(\pi-\alpha)=\sin(\beta+\gamma)\sin\alpha$$
Again,$$\sin^2\alpha=\sin\alpha\cdot\sin(\beta-\gamma)$$
Hope you can take it home from here?
A: $$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma-2 \sin \alpha \sin \beta \cos \gamma=$$
$$=\sin^2 \alpha + \sin^2 \beta -1+\cos^2(\alpha+\beta)+2 \sin \alpha \sin \beta \cos (\alpha+\beta)=$$
$$=\sin^2 \alpha-\cos^2 \beta+\cos(\alpha+\beta)\cos(\alpha-\beta)=$$
$$=\sin^2 \alpha-\cos^2 \beta+\cos^2\alpha\cos^2\beta-\sin^2\alpha\sin^2\beta=$$
$$=(1-\sin^2\beta)\sin^2\alpha-(1-\cos^2\alpha)\cos^2\beta=0.$$
A: I can say that: $\sin(\alpha+\beta)=\sin(\pi +\gamma)$.  Now, I can evaluate the expression: $$\sin(\alpha)^2+\sin(\beta)^2-\sin(\gamma)^2=\sin(\alpha)^2+\sin(\beta)^2-\sin(\alpha)^2\cos(\beta)^2-\sin(\beta)^2\cos(\alpha)^2-\sin(\alpha)\cos(\alpha)\sin(\beta)\cos(\beta)$$
Now substitute $\cos(\alpha)^2$ and $\cos(\beta)^2$, and obtain: $$2\sin(\alpha)\sin(\beta)(\sin(\alpha)\sin(\beta)-cos(\alpha)\cos(\beta))=2\sin(\alpha)\sin(\beta)(-1\cdot\cos(\alpha+\beta))=2\sin(\alpha)\sin(\beta)\cos(\gamma)$$
Also: $$\cos(\gamma)=\cos(\alpha+\beta-\pi)=\cos(\pi-(\alpha+\beta))=-\cos(\alpha+\beta)$$
From here, as you wanted: $$\sin(\alpha)^2+\sin(\beta)^2-\sin(\gamma)^2=2\sin(\alpha)\sin(\beta)\cos(\gamma)$$
