Giải bài 36 trang 207 SGK giải tích nâng cao 12

 $\begin{align}a)\,1-i\tan \dfrac{\pi }{5}&=1-i\dfrac{\sin \dfrac{\pi }{5}}{\cos \dfrac{\pi }{5}}\\ &=\dfrac{1}{\cos \dfrac{\pi }{5}}\left( \cos \dfrac{\pi }{5}-isin\dfrac{\pi }{5} \right) \\&=\dfrac{1}{\cos \dfrac{\pi }{5}}\left( \cos \dfrac{-\pi }{5}+isin\dfrac{-\pi }{5} \right) \end{align}$

$\begin{aligned} b)\,\tan \dfrac{5\pi }{8}+i&=\dfrac{sin\dfrac{5\pi }{8}}{\cos \dfrac{5\pi }{8}}+i=\dfrac{\cos \left( \dfrac{\pi }{2}-\dfrac{5\pi }{8} \right)}{\sin \left( \dfrac{\pi }{2}-\dfrac{5\pi }{8} \right)}+i=\dfrac{\cos \left( -\dfrac{\pi }{8} \right)}{\sin \left( -\dfrac{\pi }{8} \right)}+i \\ & =\dfrac{1}{\sin \left( -\dfrac{\pi }{8} \right)}\left[ \cos \left( -\dfrac{\pi }{8} \right)+i\sin \left( -\dfrac{\pi }{8} \right) \right] \\ & =\dfrac{1}{\sin \left( \dfrac{\pi }{8} \right)}\left[ \cos \left( \dfrac{7\pi }{8} \right)+i\sin \left( \dfrac{7\pi }{8} \right) \right]\\ \end{aligned}$

$\begin{aligned} c)\,1-\cos \varphi -isin\,\varphi \,&=2{{\sin }^{2}}\dfrac{\varphi }{2}-2i\sin \dfrac{\varphi }{2}\cos \dfrac{\varphi }{2} \\ & =2\sin \dfrac{\varphi }{2}\left( \sin \dfrac{\varphi }{2}-i\cos \dfrac{\varphi }{2} \right) \\ \end{aligned}$

Nếu $\sin \dfrac{\varphi }{2}>0$ thì $1-\cos \varphi -isin\,\varphi \,=2\sin \dfrac{\varphi }{2}\left[ \cos \left( \dfrac{\varphi }{2}-\dfrac{\pi }{2} \right)+i\sin \left( \dfrac{\varphi }{2}-\dfrac{\pi }{2} \right) \right]$

Nếu $\sin \dfrac{\varphi }{2}<0$ thì $1-\cos \varphi -isin\,\varphi \,=-2\sin \dfrac{\varphi }{2}\left[ \cos \left( \dfrac{\varphi }{2}+\dfrac{\pi }{2} \right)+i\sin \left( \dfrac{\varphi }{2}+\dfrac{\pi }{2} \right) \right]$.

Chú ý: Để viết dạng lượng giác của số phức $z$ ta cần làm xuất hiện dạng $z=r(\cos\varphi+i\sin\varphi)$, trong đó $r>0$.