Question

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

Answer 1

We have,

$cos\alpha +cos\beta =\frac{1}{3}andsin\alpha +sin\beta =\frac{1}{4}$

Squaring and adding, we get

$(co{s}^2\alpha +co{s}^2\beta +2cos\alpha cos\beta )+(si{n}^2\alpha +si{n}^2\beta +2sin\alpha sin\beta )=\frac{1}{9}+\frac{1}{16}\Rightarrow 1+1+2(cos\alpha +cos\beta +sin\alpha sin\beta )=\frac{25}{144}\Rightarrow 2cos(\alpha -\beta )=\frac{25}{144}-2=\frac{-263}{144}\Rightarrow cos(\alpha -\beta )=\frac{-263}{288}$

Now,

$cos\left(\frac{\alpha -\beta }{2}\right)=\sqrt{\frac{1+cos(\alpha -\beta )}{2}}=\sqrt{\frac{1-\frac{263}{288}}{2}}=\sqrt{\frac{25}{576}}=\pm \frac{5}{24}\therefore cos\left(\frac{\alpha -\beta }{2}\right)=\pm \frac{5}{24}$