Question

Let α, β be such that π < α – β < 3π. If sin α + sin β = $-\frac{21}{65}$ and cos α + cos β = $-\frac{27}{65}$, then the $\cos \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}$=
(a) $-\frac{3}{\sqrt{130}}$ (b) $\frac{3}{\sqrt{130}}$ (c) $\frac{6}{65}$ (d) $-\frac{6}{35}$

Answer 1

Dear student
$$\begin{gathered} \mathrm{Given}: \\ \mathrm{sin\alpha }+\mathrm{sin\beta }=\frac{-21}{65}\mathrm{and}\mathrm{cos\alpha }+\mathrm{cos\beta }=-\frac{27}{65} \\ \mathrm{Now}, \\ {\left(\mathrm{sin\alpha }+\mathrm{sin\beta }\right)}^2+{\left(\mathrm{cos\alpha }+\mathrm{cos\beta }\right)}^2={\left(\frac{-21}{65}\right)}^2+{\left(\frac{-27}{65}\right)}^2 \\ \Rightarrow {\sin }^2\mathrm{\alpha }+{\sin }^2\mathrm{\beta }+2\mathrm{sin\alpha }\mathrm{sin\beta }+{\cos }^2\mathrm{\alpha }+{\cos }^2\mathrm{\beta }+2\mathrm{cos\alpha }\mathrm{cos\beta }={\left(\frac{-3}{65}\right)}^2\left(7^2+9^2\right) \\ \Rightarrow 2+2\left(\mathrm{cos\alpha }\mathrm{cos\beta }+\mathrm{sin\alpha }\mathrm{sin\beta }\right)=\frac{18}{65} \\ \Rightarrow 2+2\cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)=\frac{18}{65} \\ \Rightarrow 2\cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)=\frac{-112}{65} \\ \Rightarrow \cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)=\frac{-56}{65} \\ \mathrm{Now},\cos \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)=\sqrt{\frac{\cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)+1}{2}}=\sqrt{\frac{{\displaystyle \frac{-56}{65}}+1}{2}}=\sqrt{\frac{9}{130}}=\frac{3}{\sqrt{130}} \end{gathered}$$
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