Question

Prove that:

${(cos\alpha +cos\beta )}^2+{(sin\alpha +sin\beta )}^2=4co{s}^2\left(\frac{\alpha -\beta }{2}\right)$

Answer 1

LHS

${(cos\alpha +cos\beta )}^2+{(sin\alpha +sin\beta )}^2=co{s}^2\alpha +co{s}^2\beta +2cos\alpha cos\beta +si{n}^2\alpha +si{n}^2\beta +2sin\alpha +sin\beta =(co{s}^2\alpha +si{n}^2\alpha )+(co{s}^2\beta +si{n}^2\beta )+2(cos\alpha cos\beta +sin\alpha sin\beta )=1+1+2cos(\alpha -\beta )=2+2cos(\alpha -\beta )=2(1+cos(\alpha -\beta \left)\right)=2.2co{s}^2\left(\frac{\alpha -\beta }{2}\right)=4co{s}^2\left(\frac{\alpha -\beta }{2}\right)$

= RHS