Giải bài 40 trang 209 SGK giải tích nâng cao 12

a) Ta có

$\begin{aligned} {{z}_{1}}&=\sqrt{6}-i\sqrt{2}=2\sqrt{2}\left( \dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i \right) \\ & =2\sqrt{2}\left[ \cos \left( -\dfrac{\pi }{6} \right)+i\sin \left( -\dfrac{\pi }{6} \right) \right] \\ \end{aligned}$

$\begin{aligned} {{z}_{2}}&=-2-2i=2\sqrt{2}\left( -\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}i \right) \\ & =2\sqrt{2}\left[ \cos \left( -\dfrac{3\pi }{4} \right)+i\sin \left( -\dfrac{3\pi }{4} \right) \right] \\ \end{aligned}$

$\begin{aligned} {{z}_{3}}=\dfrac{{{z}_{1}}}{{{z}_{2}}}&=\left[ \cos \left( -\dfrac{\pi }{6}+\dfrac{3\pi }{4} \right)+i\sin \left( -\dfrac{\pi }{6}+\dfrac{3\pi }{4} \right) \right] \\ & =\cos \dfrac{7\pi }{12}+i\sin \dfrac{7\pi }{12} \\ \end{aligned}$

b) Ta có

$\begin{aligned} \dfrac{{{z}_{1}}}{{{z}_{2}}}&=\dfrac{\sqrt{6}-i\sqrt{2}}{-2-2i} \\ & =\dfrac{\left( \sqrt{6}-i\sqrt{2} \right)\left( -2+2i \right)}{8} \\ & =\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{\sqrt{6}+\sqrt{2}}{4}i \\ \end{aligned}$

Mà $\begin{aligned} {{z}_{3}}=\cos \dfrac{7\pi }{12}+i\sin \dfrac{7\pi }{12} \\ \end{aligned}$

Suy ra 

$\cos \dfrac{7\pi }{12}=\dfrac{-\sqrt{6}+\sqrt{2}}{4} \\ \sin\dfrac{7\pi }{12}=\dfrac{\sqrt{6}+\sqrt{2}}{4}$