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

Prove by vector methods that sin(alpha+beta)=sin alphacos beta+cos alphasin beta

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

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

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

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

Evaluate |(cosalphacosbeta,cosalpha sinbeta , - sin alpha),(-sin beta,cosbeta,0),(sinalphacosbeta,sinalpha sinbeta,cosalpha)|

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

Show without expanding at any stage that: [0,sinalpha-cosalpha],[-sinalpha,0,sinbeta],[cosalphas-sinbeta,0]|=0

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

If alpha,beta in (-pi/2,0) such that (sin alpha+sinbeta)+(sinalpha)/(sinbeta)=0 and (sinalpha+sinbeta) (sinalpha)/(sinbeta)=-1 and lambda=lim_(n->oo) (1+(2sinalpha)^(2n))/((2sinbeta)^(2n)) then :