cos(A+B)=cosAcosB-sinAsinB (trig identity).
cos(5(pi)/6)=-sqrt(3)/2; sin(5(pi)/6)=1/2. Putting A=5(pi)/6 and B=isin(5(pi)/6)=i/2:
cos(A+B)=(-sqrt(3)/2)(cos(i/2))-(1/2)sin(i/2).
Also, by de Moivre: e^(ix)=cosx+isinx; e^-(ix)=cosx-isinx,
so e^(ix)+e^-(ix)=2cosx and e^(ix)-e^-(ix)=2isinx.
2cos(ix)=e^-x+e^x; 2isin(ix)=e^-x-e^x; -2sin(ix)=i(e^-x-e^x).
cos(i/2)=(e^-(1/2)+e^(1/2))/2; sin(i/2)=(-i(e^-(1/2)-e^(1/2))/2=i(e^(1/2)-e^-(1/2))/2.
[sqrt(3)(cos(A+B)]^4=9[(-sqrt(3)/2)(e^-(1/2)+e^(1/2))/2-(1/2)i(e^(1/2)-e^-(1/2))/2]^4.
Now, this answer is clearly very complicated and we still have to calculate the fourth power. In order to assess whether this is leading to the expected answer provided, we need to work out the expression in square brackets as far as possible. I believe that the original question may have been wrongly entered, because of the complexity, and should have been [sqrt(3)(cos(5pi/6)+isin(5pi/6)]^4, which would result in an easier application of de Moivre. Proceeding briefly on this assumption we get:
9e^(i5(pi)/6)^4=9(cos(20(pi)/6)+isin(20(pi)/6) applying e^(iX)=cosX+isinX, so that e^(niX)=cos(nX)+isin(nX), where n=4 and X=5(pi)/6.
20(pi)/6=10(pi)/3=2(pi)+4(pi)/3, in the third quadrant. cos(4(pi)/3)=-1/2 and sin(4(pi)/3)=-sqrt(3)/2.
The expression becomes: 9(-1/2-isqrt(3)/2) or -9/2-i9sqrt(3)/2 in standard form (a+ib), which is the provided answer, proving that the question was misstated. Both the real and imaginary parts are negative, caused by multiplication of the angle taking it into the third quadrant where sine and cosine are negative.