Cho góc α 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:

Vì $\frac{\pi }{2} < \alpha < \pi$ nên sin α > 0. Mặt khác từ sin² α + cos² α = 1 suy ra

$\sin \alpha = \sqrt {1 - {{\cos }^2}\alpha } = \sqrt {1 - {{\left( { - \frac{1}{{\sqrt 3 }}} \right)}^2}} = \frac{{\sqrt 6 }}{3}$.

a) $\sin \left( {\alpha + \frac{\pi }{6}} \right)$$= \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6}$

$= \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} + \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{1}{2} = \frac{{3\sqrt 2 - \sqrt 3 }}{6}$.

b) $\cos \left( {\alpha + \frac{\pi }{6}} \right)$$= \cos \alpha \cos \frac{\pi }{6} - \sin \alpha \sin \frac{\pi }{6}$

$= - \frac{1}{{\sqrt 3 }}.\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 - \sqrt 6 }}{6}$.

c) $\sin \left( {\alpha - \frac{\pi }{3}} \right)$$= \sin \alpha \cos \frac{\pi }{3} - \cos \alpha \sin \frac{\pi }{3}$

$= \frac{{\sqrt 6 }}{3}.\frac{1}{2} - \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} = \frac{{\sqrt 6 + 3}}{6}$.

d) $\cos \left( {\alpha - \frac{\pi }{6}} \right)$$= \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6}$

$= - \frac{1}{{\sqrt 3 }}.\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}$.