Question

6sin15(градусов)*cos15(градусов)/2cos^2*15(градусов)-1

Answer 1

$\frac{6sin15^0\cdot cos15^0}{2cos^215^0-1}=\frac{3sin30^0}{cos30^0} =3tg30^0=3\cdot \frac{1}{\sqrt3}=\sqrt3\\\\Formyla:\; \; cos^2 \alpha =\frac{1+cos2 \alpha }{2}\; \; \to \; \; cos2 \alpha =2cos^2 \alpha -1\\\\ili\; \; \; cos2 \alpha =cos^2 \alpha -sin^2 \alpha =cos^2 \alpha -(1-cos^2 \alpha )=2cos^2 \alpha -1$