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

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

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

Show by vector method that sin(alpha-beta)=sinalphacosbeta-cosalpha sinbeta.

Prove that |cosalpha-cosbeta|lt=|alpha-beta|

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

f(alpha,beta) = cos^2(alpha)+ cos^2(alpha+beta)- 2 cosalpha cosbeta cos(alpha+beta) is

If cosalpha+cosbeta+cosgamma=0=sinalpha+sinbeta+singamma , then which of the following is/are true:- (a) cos(alpha-beta)+cos(beta-gamma)+cos(gamma-delta)=-3/2 (b) cos(alpha-beta)+cos(beta-gamma)+cos(gamma-delta)=-1/2 (c) sumcos2alpha+2cos(alpha+beta)+2cos(beta+gamma)+2cos(gamma+alpha)=0 (d) sumsin2alpha+2sin(alpha+beta)+2sin(beta+gamma)+2sin(gamma+alpha)=0

Prove that: (cosalpha+cosbeta)^2+(sinalpha+sinbeta)^2=4cos^2((alpha-beta)/2)

Prove that: (cosalpha-cosbeta)^2+(sinalpha-sinbeta)^2=4sin^2((alpha-beta)/2)^(\ )

Show that: sin^2 alpha + sin^2 beta + 2sinalpha sinbeta cos(alpha+beta)=sin^2 (alpha+beta)