Question

If $\cos \alpha +\cos \beta =\frac{1}{3}$ and sin $\sin \alpha +\sin \beta =\frac{1}{4}$, prove that $\cos \frac{\alpha -\beta }{2}=\pm \frac{5}{24}$

Answer 1

Squaring and adding equations $\cos \mathrm{\alpha }+\cos \mathrm{\beta }=\frac{1}{3}$ and $\sin \alpha +\sin \beta =\frac{1}{4}$, we get

$$\begin{gathered} \left({\cos }^2\mathrm{\alpha }+{\cos }^2\mathrm{\beta }+2\mathrm{cos\alpha }\times \mathrm{cos\beta }\right)+\left({\sin }^2\mathrm{\alpha }+{\sin }^2\mathrm{\beta }+2\mathrm{sin\alpha }\times \mathrm{sin\beta }\right)=\frac{1}{9}+\frac{1}{16} \\ \Rightarrow 1+1+2\left(\mathrm{cos\alpha }\times \mathrm{cos\beta }+\mathrm{sin\alpha }\times \mathrm{sin\beta }\right)=\frac{25}{144} \\ \Rightarrow 2+2\cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)=\frac{25}{144}\left(\because \cos \left(A-B\right)=\cos A\times \cos B+\sin A\times \sin B\right) \\ \Rightarrow \cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)=-\frac{263}{288}...\left(1\right) \end{gathered}$$

Now,

$$\begin{gathered} {\cos }^2\left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)=\frac{1+\cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)}{2} \\ =\frac{1-{\displaystyle \frac{263}{288}}}{2}[\mathrm{From}(1\left)\right] \\ =\frac{25}{576} \\ =\pm \frac{5}{24} \end{gathered}$$