How do you prove that sqrt(3)cos(x+pi/6) - cos(x+pi/3) = cos(x)-sqrt3sinx ?

1 Answer
May 2, 2018

LHS=sqrt3cos(x+pi/6)-cos(x-pi/3)

=sqrt3[cosx*cos(pi/6)-sinx*sin(pi/6)]-[cosx*cos(pi/3)-sinx*sin(pi/3)]

=sqrt3[cosx*(sqrt3/2)-sinx*(1/2)]-[cosx*(1/2)-sinx*(sqrt3/2)]

=(3cosx-sqrt3sinx)/2-(cosx-sqrt3sinx)/2

=(3cosx-sqrt3sinx-cosx+sqrt3sinx)/2

=(2cosx)/2=cosx=RHS