IIT-JEE 1980Question 1
Given $\alpha+\beta-\gamma=\pi$, prove that $\sin^{2}\alpha+\sin^{2}\beta-\sin^{2}\gamma=2\sin\alpha\sin\beta\cos\gamma$.
Trigonometric Ratio & Identities
Answerproved
Solution
Concept. $\gamma=\alpha+\beta-\pi$; use $\sin^2P-\sin^2Q=\sin(P-Q)\sin(P+Q)$ and then sum-to-product. Formulas used. (1) $\sin^2P-\sin^2Q=\sin(P-Q)\sin(P+Q)$(2) $\sin P-\sin Q=2\cos\frac{P+Q}{2}\sin\frac{P-Q}{2}$Step 1 - Rewrite $\gamma$. $\sin^2\gamma=\sin^2(\alpha+\beta)$ and $\cos\gamma=-\cos(\alpha+\beta)$. Step 2 - Use (1). $\sin^2\alpha-\sin^2(\alpha+\beta)=\sin(-\beta)\sin(2\alpha+\beta)$, so $$\text{LHS}=\sin\beta\left[\sin\beta-\sin(2\alpha+\beta)\right].$$ Step 3 - Use (2). $\sin\beta-\sin(2\alpha+\beta)=-2\cos(\alpha+\beta)\sin\alpha$, so $$\text{LHS}=-2\sin\alpha\sin\beta\cos(\alpha+\beta)=2\sin\alpha\sin\beta\cos\gamma.$$ Hence proved.