[latex] \sqrt{75} - \sqrt{300} sin^2 \frac{13Pi}{12} [/latex]

[latex]sin^2 \alpha =\frac{1-cos2 \alpha }{2}\; \to \\\\sin^2\frac{13\pi}{12}=sin^2(\pi +\frac{\pi}{12})=sin^2\frac{\pi}{12}=\frac{1-cos\frac{\pi}{6}}{2}=\frac{1-\frac{\sqrt3}{2}}{2}=\frac{2-\sqrt3}{4}\\\\\sqrt{75}-\sqrt{300}sin^2\frac{13\pi }{12}=\sqrt{75}(1-2\cdot \frac{2-\sqrt3}{4})=5\sqrt3\cdot \frac{\sqrt3}{2}=\frac{15}{2}=7,5[/latex]