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

\[\cos \alpha + \cos \beta = \frac{1}{3}\]  and \[\sin\alpha + \sin \beta = \frac{1}{4}\] , we get

\[\left( \cos^2 \alpha + \cos^2 \beta + 2cos\alpha \times cos\beta \right) + \left( \sin^2 \alpha + \sin^2 \beta + 2sin\alpha \times sin\beta \right) = \frac{1}{9} + \frac{1}{16}\]
\[ \Rightarrow 1 + 1 + 2\left( cos\alpha \times cos\beta + sin\alpha \times sin\beta \right) = \frac{25}{144}\]
\[ \Rightarrow 2 + 2\cos\left( \alpha - \beta \right) = \frac{25}{144} \left( \because \cos\left( A - B \right) = \text{ cos } A \times \text{ cos }B + \text{ sin } A \times \text{ sin } B \right)\]
\[ \Rightarrow \cos\left( \alpha - \beta \right) = - \frac{263}{288} . . . (1)\]

Now,

\[\cos^2 \left( \frac{\alpha - \beta}{2} \right) = \frac{1 + \cos\left( \alpha - \beta \right)}{2}\]
\[ = \frac{1 - \frac{263}{288}}{2} [\text{ From }  (1)]\]
\[ = \frac{25}{576}\]
\[ = \pm \frac{5}{24}\]