Bài 1.31 trang 41 SGK Toán 11 tập 1 - Kết nối tri thức

Đề bài

Cho góc \(\alpha \) thỏa mãn \(\frac{\pi }{2} < \alpha  < \pi ,\cos \alpha  =  - \frac{1}{{\sqrt 3 }}\). Tính giá trị của các biểu thức sau:

a) \(\sin \left( {\alpha  + \frac{\pi }{6}} \right)\);     

b) \(\cos \left( {\alpha  + \frac{\pi }{6}} \right);\)           

c) \(\sin \left( {\alpha  - \frac{\pi }{3}} \right)\);      

d) \(\cos \left( {\alpha  - \frac{\pi }{6}} \right)\).

Lời giải chi tiết

\(\begin{array}{l}{\sin ^2}\left( \alpha  \right) + {\cos ^2}\left( \alpha  \right) = 1 \Leftrightarrow {\sin ^2}\left( \alpha  \right) + {\left( { - \frac{1}{{\sqrt 3 }}} \right)^2} = 1\\ \Leftrightarrow {\sin ^2}\left( \alpha  \right) + \frac{1}{3} = 1 \Leftrightarrow {\sin ^2}\left( \alpha  \right) = \frac{2}{3}\\ \Leftrightarrow \sin \left( \alpha  \right) = \sqrt {\frac{2}{3}}  = \frac{{\sqrt 6 }}{3}\end{array}\)

Ta có:

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

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

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

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