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?
Best Answer
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?