Bài 1.12 trang 19

Đề bài

Biết $\cos \alpha =-\frac{1}{5}$ và $\pi <\alpha <\frac{3\pi }{2}$, tính:

a) $\cos \left(\alpha -\frac{\pi }{3}\right);$

b) $\tan \left(\alpha +\frac{\pi }{4}\right);$

c) $\sin \frac{\alpha }{2}.$

Lời giải chi tiết

a) ${\sin }^2\alpha =1-{\cos }^2\alpha =\frac{24}{25}$

$\Rightarrow \sin \alpha =-\frac{2\sqrt{6}}{5}$ (Vì $\pi <\alpha <\frac{3\pi }{2}$)

$\cos \left(\alpha -\frac{\pi }{3}\right)=\cos \alpha \cos \frac{\pi }{3}+\sin \alpha \sin \frac{\pi }{3}=-\frac{1+6\sqrt{2}}{10}$

b) $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }=2\sqrt{6}$

$\tan \left(\alpha +\frac{\pi }{4}\right)=\frac{\tan \alpha +\tan \frac{\pi }{4}}{1-\tan \alpha \tan \frac{\pi }{4}}=-\frac{25+4\sqrt{6}}{23}$

c) ${\sin }^2\frac{\alpha }{2}=\frac{1-\cos \alpha }{2}=\frac{3}{5}\Rightarrow \sin \frac{\alpha }{2}=\frac{\sqrt{15}}{5}$