Question

If $\sin \left(\alpha \right)+\sin \left(\beta \right)=a$ and $\cos \left(\alpha \right)+\cos \left(\beta \right)=b$, then prove that $\cos \left(\alpha -\beta \right)=\frac{\left[a^2+b^2-2\right]}{2}$

Answer 1

Prove the given condition:

Given:

$\sin \left(\alpha \right)+\sin \left(\beta \right)=a\to \left(1\right)$

$\cos \left(\alpha \right)+\cos \left(\beta \right)=b\to \left(2\right)$

On squaring and adding $\left(1\right)$ and $\left(2\right)$, we have,

$\begin{array}{l}{\sin }^2\left(\alpha \right)+{\sin }^2\left(\beta \right)+2\times \sin \left(\alpha \right)\sin \left(\beta \right)+{\cos }^2\left(\alpha \right)+{\cos }^2\left(\beta \right)+2\times \cos \left(\alpha \right)\cos \left(\beta \right)=a^2+b^2\\ \Rightarrow 1+1+2(\sin \left(\alpha \right)\sin \left(\beta \right)+\cos \left(\alpha \right)\cos \left(\beta \right))=a^2+b^2\\ \Rightarrow 2(\sin \left(\alpha \right)\sin \left(\beta \right)+\cos \left(\alpha \right)\cos \left(\beta \right))=a^2+b^2-2\\ \Rightarrow 2\cos (\alpha -\beta )=a^2+b^2-2[\because \cos (A-B)=\sin \left(A\right)\sin \left(B\right)+\cos \left(A\right)\cos \left(B\right)]\\ \Rightarrow \cos (\alpha -\beta )=\frac{a^2+b^2-2}{2}\end{array}$

Hence proved that $\cos \left(\alpha -\beta \right)=\frac{\left[a^2+b^2-2\right]}{2}$.