$\displaystyle L=\int\limits^{-\frac{\pi}{6}}_{-\frac{\pi}{3}}\sqrt{\sin^2\phi+(1-\cos \phi)^2}d\phi=\int\limits^{-\frac{\pi}{6}}_{-\frac{\pi}{3}}\sqrt{\sin^2\phi+1-2\cos\phi+\cos^2\phi}d\phi=\\ \\ \\ =\int\limits^{-\frac{\pi}{6}}_{-\frac{\pi}{3}}\sqrt{\sin^2\phi+1-2\cos\phi+1-\sin^2\phi}\, d\phi=\int\limits^{-\frac{\pi}{6}}_{-\frac{\pi}{3}}\sqrt{2-2\cos\phi}\, d\phi=\\ \\ \\$

$\displaystyle=\int\limits^{-\frac{\pi}{6}}_{-\frac{\pi}{3}}\sqrt{2(1-\cos \phi)}\, d\phi=\int\limits^{-\frac{\pi}{6}}_{-\frac{\pi}{3}}\sqrt{4\sin^2\frac{\phi}{2}}\, d\phi=2\int\limits^{-\frac{\pi}{6}}_{-\frac{\pi}{3}}\bigg|\sin\frac{\phi}{2}\bigg|~ d\phi=\\ \\ =-2\cdot 2 \left(-\cos\frac{\phi}{2}\right)\bigg|^{-\frac{\pi}{6}}_{-\frac{\pi}{3}}=4\cos\frac{\pi}{12}-4\cos\frac{\pi}{6}=4\cos\dfrac{\pi}{12}-4\cdot\dfrac{\sqrt{3}}{2}=\\ \\ \\ =4\cos\dfrac{\pi}{12}-2\sqrt{3}~~\boxed{=}$

распишем cos п/12:

$\cos \dfrac{\pi}{12}=\cos \left(\dfrac{\pi}{4}-\dfrac{\pi}{6}\right)=\cos\dfrac{\pi}{4}\cos\dfrac{\pi}{6}+\sin\dfrac{\pi}{4}\sin\dfrac{\pi}{6}=\dfrac{\sqrt{2}}{2}\cdot \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}\cdot \dfrac{1}{2}=\\ \\ \\ =\dfrac{\sqrt{6}+\sqrt{2}}{4}$

Окончательно получим ответ

$\boxed{=}~ 4\cdot \dfrac{\sqrt{6}+\sqrt{2}}{4}-2\sqrt{3}=\sqrt{6}+\sqrt{2}-2\sqrt{3}$

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4 года назад

Помогите решить! Sin2x=sinx+cosx

Сергей Зорькин [7]

Решение смотрите в приложении

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3 года назад