tex]...

October 2021 1 17 Report

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

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NNNLLL54

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

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