Sum and Difference Identities?

The answer is as follows:

Put (5pi/6) = y. Now cos(x+y)  = cosx*cosy - sinx*siny and (5pi/6) = pi - pi/6.
Then cosy = cos(pi-pi/6) = -cos(pi/6) = -(1/2)rt3;
and siny = sin(pi-pi/6) = sin(pi/6) = (1/2).
Then cos[x+(5pi/6)] = -(1/2)rt3*cosx - (1/2)*sinx = - (1/2)[rt3*cosx + sinx].

cos (x + 5π/6)
= -cos(π/6 - x)
= -sin(x)/2 - 1/2 sqrt(3) cos(x)

cos(x+5pi/6)=
cos(x)cos(5pi/6)-sin(x)sin(5pi/6)=
-cos(x)cos(pi/6)-sin(x)sin(pi/6)=
-[sqr(3)cos(x)+sin(x)]/2

cos(A + B) => cosAcosB - sinAsinB

so, cos(x + 5π/6) => cos(x)cos(5π/6) - sin(x)sin(5π/6)

Now, cos(5π/6) = -√3/2 and sin(5π/6) = 1/2

Hence, cos(x + 5π/6) => (-√3/2)cosx - (1/2)sinx

i.e. (-1/2)(√3cosx + sinx)

:)>   

(-√3/2)cosx - (1/2)sinx

stop doing this to us please