Помогите с Б7. Мое домашнее задание во вложениях.

Решите задачу:

$\sqrt{108}cos^2\frac{11 \Pi}{12} - \sqrt{27} = \\ =\sqrt{36*3}cos^2( \Pi - \frac{\Pi}{12}) - \sqrt{9*3} = \\ =6 \sqrt{3}\ [-cos( \frac{\Pi}{12})]^2 - 3\sqrt{3} = \\ =3\sqrt{3}*[2cos^2( \frac{\Pi}{12}) - 1]=... \ \ \ \ \\ ( {\mathrm{formula \ cosinusa \ dvojnogo \ ugla: \\ } \\ \ \ \bold {cos(2x) =2cos^2(x) - 1}}) \\ ... =3\sqrt{3}*[2cos^2( \frac{\Pi}{12}) - 1]= \\ =3\sqrt{3}*cos( \frac{2\Pi}{12}) =3\sqrt{3}*cos( \frac{\Pi}{6})= \\ =3\sqrt3* \frac{\sqrt3}{2} = \frac{3*3}{2} = 4 \frac12 \\$