Jak udowodnić, że sqrt (3) cos (x + pi / 6) - cos (x + pi / 3) = cos (x) -sqrt3sinx?

# 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 #