Question

Помогите с Б7. Мое домашнее задание во вложениях.

Answer 1

3sqrt(3)(2cos^2(pi/12)-1)=3sqrt(3)cospi/6=3*sqrt(3)*sqrt(3)/2=4,5

Answer 2

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