Question

If $\sin \alpha +\sin \beta =a\mathrm{and}\cos \alpha +\cos \beta =b$, prove that

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

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

Answer 1

The given equations are $\sin \alpha +\sin \beta =a\mathrm{and}\cos \alpha +\cos \beta =b$.

(i)

$$\begin{gathered} \because \sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2} \\ \therefore 2\sin \frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\cos \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}=a...\left(1\right) \end{gathered}$$

Now, using the identity $\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}$ for the LHS of$\cos \alpha +\cos \beta =b$, we get

$2\cos \frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\cos \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}=b...\left(2\right)$

On dividing (1) by (2), we get

$\tan \frac{\mathrm{\alpha }+\mathrm{\beta }}{2}=\frac{a}{b}$

We know,

$\mathrm{sin\theta }=\frac{2\tan {\displaystyle \frac{\mathrm{\theta }}{2}}}{1+{\tan }^2{\displaystyle \frac{\mathrm{\theta }}{2}}}$
$$\begin{gathered} \therefore \sin \left(\mathrm{\alpha }+\mathrm{\beta }\right)=\frac{2\tan \left({\displaystyle \frac{\mathrm{\alpha }+\mathrm{\beta }}{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{\mathrm{\alpha }+\mathrm{\beta }}{2}}\right)} \\ \Rightarrow \sin \left(\mathrm{\alpha }+\mathrm{\beta }\right)=\frac{2\times {\displaystyle \frac{a}{b}}}{1+{\displaystyle \frac{a^2}{b^2}}}=\frac{2ab}{a^2+b^2} \end{gathered}$$

(ii)

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