Question

If \[\sin \alpha + \sin \beta = a \text{ and }  \cos \alpha + \cos \beta = b\] , prove that

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

 

Answer 1

The given equations are \[\sin \alpha + \sin \beta = a \text{ and }  \cos \alpha + \cos \beta = b\]

\[\text{ On squaring } sin\alpha + sin\beta = \text{ a and }  cos\alpha + cos\beta = \text{ b and adding them, we get} \]
\[ \sin^2 \alpha + \sin^2 \beta + 2 \times sin\alpha sin\beta + \cos^2 \alpha + \cos^2 \beta + 2 \times cos\alpha cos\beta = a^2 + b^2 \]
\[ \Rightarrow 1 + 1 + 2\left( sin\alpha sin\beta + cos\alpha cos\beta \right) = a^2 + b^2 \]
\[ \Rightarrow 2\left( sin\alpha sin\beta + cos\alpha cos\beta \right) = a^2 + b^2 - 2\]
\[ \Rightarrow 2\cos\left( \alpha - \beta \right) = a^2 + b^2 - 2 \left( \because \cos\left( A - B \right) = sinAsinB + cosAcosB \right)\]
\[ \Rightarrow \cos\left( \alpha - \beta \right) = \frac{a^2 + b^2 - 2}{2}\]