Вычислите [latex] sin^{4} \frac{ \pi }{12} + cos^{4} \frac{ \pi }{12}[/latex]

1-сosπ/6)²/4+(1+cosπ/6)²/4=(1-2cosπ/6+cos²π/6+1+2cosπ/6+cos²π/6)/4= =(2+2cos²π/6)/4=(1+cos²π/6)/2=(1+3/4)/2=7/8

[latex]\sin^4\frac{\pi}{12}+\cos^4\frac{\pi}{12}=(\sin^2\frac{\pi}{12})^2+(\cos^2\frac{\pi}{12})^2=[/latex] [latex]=(\frac{1-\cos\frac{2\pi}{12}}{2})^2+(\frac{1+\cos\frac{2\pi}{12}}{2})^2=\frac{(1-\cos\frac{\pi}{6})^2}{4}+\frac{(1+\cos\frac{\pi}{6})^2}{4}=\frac{(1-\frac{\sqrt3}{2})^2+(1+\frac{\sqrt3}{2})^2}{4}=\\\\=\frac{1-\sqrt3+\frac{3}{4}+(1+\sqrt3+\frac{3}{4})}{4}=\frac{1-\sqrt3+\frac{3}{4}+1+\sqrt3+\frac{3}{4}}{4}=\frac{2+\frac{6}{4}}{4}=\frac{\frac{14}{4}}{4}=\frac{14}{16}=\frac{7}{8}.[/latex]