Prove by vector metod the following formula of plane trigonometry `cos(alpha-beta)=cosalpha cosbeta+sinalpha sinbeta`

Using vectors, prove that cos(alpha-beta)=cosalpha cosbeta+sinalphasinbeta

(cosalpha+cosbeta)^2+(sinalpha+sinbeta)^2 =

sinalpha+sinbeta=a ,cosalpha+cosbeta=b=>sin(alpha+beta)

Simplify (sinalpha+sinbeta)/(cosalpha-cosbeta)+(cosalpha+cosbeta)/(sinalpha-sinbeta)

If sinalpha+sinbeta=a ,cosalpha+cosbeta=b=>sin(alpha+beta) =

In alpha ge 0 ,betage0 and alpha+betalepi , then ii) cos(alpha-beta) = cos alpha cosbeta+sinalpha sinbeta

In alpha ge 0 ,betage0 and alpha+betalepi , then i) cos(alpha+beta) = cos alpha cosbeta-sinalpha sinbeta

The expression (sinalpha+sinbeta)/(cosalpha+cosbeta) is equal to

If cosalpha+cosbeta=0=sinalpha+sinbeta" then "cos2alpha+cos2beta

By geometrical interpretation, prove that (i) sin(alpha+beta)=sin alpha cos beta+sinbeta cosalpha (ii) cos(alpha+beta)=cosalpha cosbeta -sin alpha sinbeta