Question

Prove that, ${(\cos \alpha +\cos \beta )}^2+{(\sin \alpha +\sin \beta )}^2=4{\cos }^2\left(\frac{\alpha -\beta }{2}\right)$

Answer 1

We have,

${(\cos \alpha +\cos \beta )}^2+{(\sin \alpha +\sin \beta )}^2$

$\Rightarrow {\cos }^2\alpha +{\cos }^2\beta +2\cos \alpha \cos \beta +{\sin }^2\alpha +{\sin }^2\beta +2\sin \alpha \sin \beta$

$\Rightarrow {\sin }^2\alpha +{\cos }^2\alpha +{\sin }^2\beta +{\cos }^2\beta +2\sin \alpha \sin \beta +2\cos \alpha \cos \beta$

$\Rightarrow 1+1+2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )(\therefore {\sin }^2\alpha +{\cos }^2\alpha =1)$

$\Rightarrow 2+2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )$

$\Rightarrow 2+2\left(\cos (\alpha -\beta )\right)$

$\Rightarrow 2(1+\cos (\alpha -\beta ))\therefore \cos \theta =2{\cos }^2\frac{\theta }{2}-1$

$\Rightarrow 2(1+2{\cos }^2\frac{(\alpha -\beta )}{2}-1)$

$\Rightarrow 4{\cos }^2\frac{(\alpha -\beta )}{2}$

Hence, this is the answer.