a) Tính \(\sin {{25\pi } \over 6} + \cos {{25\pi } \over 3} + \tan ( - {{25\pi } \over 4})\)
b) Biết \(\sin (\pi + \alpha ) = - {1 \over 3}\) , hãy tính \(\cos (2π – α)\) và \(\sin ({{3\pi } \over 2} - \alpha )\)
Đáp án
a) Ta có:
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\(\eqalign{
& \sin {{25\pi } \over 6} = \sin (4\pi + {\pi \over 6}) = \sin {\pi \over 6} = {1 \over 2} \cr
& \cos {{25\pi } \over 3} = \cos (8\pi + {\pi \over 3}) = \cos {\pi \over 3} = {1 \over 2} \cr
& \tan ( - {{25\pi } \over 4}) = - tan(6\pi + {\pi \over 4}) = - \tan {\pi \over 4} = - 1 \cr
& \Rightarrow \sin {{25\pi } \over 6} + \cos {{25\pi } \over 3} + \tan ( - {{25\pi } \over 4}) = 0 \cr} \)
b) Ta có:
\(\eqalign{
& \sin (\pi + \alpha ) = - {1 \over 3} \Rightarrow \sin \alpha = {1 \over 3} \cr
& \cos (2\pi - \alpha ) = \cos ( - \alpha ) = \cos \alpha = \pm \sqrt {1 - {{\sin }^2}\alpha } \cr&= \pm {{2\sqrt 2 } \over 3} \cr
& \tan (\alpha - 7\pi ) = \tan \alpha = {{\sin \alpha } \over {\cos \alpha }} = \pm {1 \over {2\sqrt 2 }} \cr
& \sin ({{3\pi } \over 2} - \alpha ) = \sin (\pi + {\pi \over 2} - \alpha ) = - \sin ({\pi \over 2} - \alpha )\cr& = - \cos \alpha= \pm {{2\sqrt 2 } \over 3} \cr} \)