Cho sin α = 2/3 , pi/ 2 < α < pi . Khi đó: a) cos α = − 5/3

a) Đúng

b) Đúng

c) Sai

d) Sai

 

\(\sin \alpha = \frac{2}{3},\frac{\pi }{2} < \alpha < \pi \)

Ta có: \({\cos ^2}\alpha = 1 - {\sin ^2}\alpha = 1 - {\left( {\frac{2}{3}} \right)^2} = \frac{5}{9} \Rightarrow \cos \alpha = \pm \frac{{\sqrt 5 }}{3}\)

Vì \(\frac{\pi }{2} < \alpha < \pi \) nên \(\cos \alpha = - \frac{{\sqrt 5 }}{3}\)

\(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = - \frac{{2\sqrt 5 }}{5}\).

\(\cos \left( {\frac{\pi }{3} + \alpha } \right) = \cos \frac{\pi }{3}\cos \alpha - \sin \frac{\pi }{3}\sin \alpha = \frac{1}{2} \cdot \left( {\frac{{ - \sqrt 5 }}{3}} \right) - \frac{{\sqrt 3 }}{2} \cdot \frac{2}{3} = \frac{{ - \sqrt 5 - 2\sqrt 3 }}{6}\)

cosπ4−α=cosπ4cosα+sinπ4sinα=22⋅−53+22⋅23=−10+226