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?

$$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma-2 \sin \alpha \sin \beta \cos \gamma=$$
$$=\sin^2 \alpha + \sin^2 \beta -1+\cos^2(\alpha+\beta)+2 \sin \alpha \sin \beta \cos (\alpha+\beta)=$$
$$=\sin^2 \alpha-\cos^2 \beta+\cos(\alpha+\beta)\cos(\alpha-\beta)=$$
$$=\sin^2 \alpha-\cos^2 \beta+\cos^2\alpha\cos^2\beta-\sin^2\alpha\sin^2\beta=$$
$$=(1-\sin^2\beta)\sin^2\alpha-(1-\cos^2\alpha)\cos^2\beta=0.$$
