Question

If sin α + sin β = a and cos α + cos β = b, show that

\[\sin \left( \alpha + \beta \right) = \frac{2ab}{a^2 + b^2}\]

 

Answer 1

\[a^2 + b^2 = \left( \sin\alpha + \sin\beta \right)^2 + (\cos\alpha + \cos\beta)^2 \]

\[ \Rightarrow a^2 + b^2 = \sin^2 \alpha + \sin^2 \beta + 2\sin\alpha\sin\beta + \cos^2 \alpha + \cos^2 \beta + 2\cos\alpha\cos\beta\]

\[ \Rightarrow a^2 + b^2 = \sin^2 \alpha + \cos^2 \alpha + \sin^2 \beta + \cos^2 \beta + 2\left( \sin\alpha\sin\beta + \cos\alpha\cos\beta \right)\]

\[ \Rightarrow a^2 + b^2 = 2 + 2 \cos(\alpha - \beta) . . . (1)\]
Now,
\[b^2 - a^2 = {(\cos\alpha + \cos\beta)}^2 - \left( \sin\alpha + \sin\beta \right)^2 \]
\[ \Rightarrow b^2 - a^2 = \cos^2 \alpha + \cos^2 \beta - \sin^2 \alpha - \sin^2 \beta + 2\cos\alpha\cos\beta - 2\sin\alpha\sin\beta\]
\[ \Rightarrow b^2 - a^2 = ( \cos^2 \alpha - \sin^2 \beta) + ( \cos^2 \beta - \sin^2 \alpha) - 2\cos(\alpha + \beta)\]
\[ \Rightarrow b^2 - a^2 = 2\cos(\alpha + \beta)\cos(\alpha - \beta) + 2\cos(\alpha - \beta)\]
\[ \Rightarrow b^2 - a^2 = \cos(\alpha + \beta)(2 + 2 \cos(\alpha - \beta)) . . . (2)\]
From (1) and (2), we have

\[b^2 - a^2 = \cos(\alpha + \beta)\left( a^2 + b^2 \right) \]

\[ \Rightarrow \frac{b^2 - a^2}{a^2 + b^2} = \cos(\alpha + \beta)\]
\[\Rightarrow \sin\left( \alpha + \beta \right) = \sqrt{1 - \cos^2 (\alpha + \beta)}\]
\[ \Rightarrow \sin\left( \alpha + \beta \right) = \sqrt{1 - \left( \frac{b^2 - a^2}{b^2 + a^2} \right)^2} = \sqrt{\frac{b^4 + a^4 - b^4 - a^4 + 4 a^2 b^2}{\left( b^2 + a^2 \right)^2}}\]
\[ \Rightarrow \sin\left( \alpha + \beta \right) = \frac{2ab}{a^2 + b^2}\]