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