Question

Let $\alpha ,\beta$be such that $\pi <(\alpha -\beta )<3\pi$. If $\sin \alpha +\sin \beta =\frac{-21}{65}$ and $cos\alpha +\cos \beta =\frac{-27}{65}$, then the value of$\frac{\cos (\alpha -\beta )}{2}$ is:

• A. $\frac{-6}{65}$
• B. $\frac{3}{\sqrt{130}}$
• C. $\frac{6}{65}$
• D. $-\frac{3}{\sqrt{130}}$

Answer 1

The correct option is D

$-\frac{3}{\sqrt{130}}$

Explanation for the correct option:

Step 1: Simplifying the given equation

The given equations are

$$\begin{gathered} \sin \alpha +\sin \beta =-\frac{21}{65}\dots .\left(i\right) \\ \cos \alpha +\cos \beta =-\frac{21}{65}\dots .\left(ii\right) \end{gathered}$$

Squaring equations $\left(i\right)$ and $\left(ii\right)$, we get

$$\begin{gathered} {\left(\sin \alpha +\sin \beta \right)}^2={\left(-\frac{21}{65}\right)}^2\dots \dots \left(iii\right) \\ {\left(\cos \alpha +\cos \beta \right)}^2={\left(-\frac{21}{65}\right)}^2\dots \dots \left(iv\right) \end{gathered}$$

Step 2: Finding the value of$\frac{\cos (\alpha -\beta )}{2}$

Adding equations $\left(iii\right)$ and $\left(iv\right)$

$$\begin{gathered} \Rightarrow \left({\sin }^2\alpha +{\sin }^2\beta +2\sin \alpha \sin \beta \right)+\left({\cos }^2\alpha +{\cos }^2\beta +2\cos \alpha \cos \beta \right)={\left(\frac{21}{65}\right)}^2+{\left(\frac{27}{65}\right)}^2 \\ \Rightarrow ({\sin }^2\alpha +{\cos }^2\alpha )+({\sin }^2\beta +{\cos }^2\beta )+2(\sin \alpha \sin \beta +\cos \alpha \cos \beta )=\frac{\left[{\left(21\right)}^2+{\left(27\right)}^2\right]}{{\left(65\right)}^2} \\ \Rightarrow 1+1+2\cos (\alpha –\beta )=\frac{18}{65} \\ \Rightarrow 2\left[1+\cos (\alpha –\beta )\right]=\frac{18}{65}\left[\because 1+\cos 2A=2{\cos }^{2}A,\right] \\ \Rightarrow 2\left[2{\cos }^2\frac{(\alpha –\beta )}{2}\right]=\frac{18}{65} \\ \Rightarrow {\cos }^2\frac{(\alpha –\beta )}{2}=\frac{9}{130} \\ \Rightarrow \cos \frac{(\alpha –\beta )}{2}=\pm \sqrt{\frac{9}{130}} \\ \Rightarrow \cos \frac{(\alpha –\beta )}{2}=\pm \sqrt{\frac{9}{130}} \\ \Rightarrow \cos \frac{(\alpha –\beta )}{2}=-\frac{3}{\sqrt{130}}\left[\because \text{given}\pi <(\alpha –\beta )<3\pi \right] \end{gathered}$$Therefore, option (D) is the correct answer.