Bài 1.31 trang 41

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

${\sin }^2\left(\alpha \right)+{\cos }^2\left(\alpha \right)=1$

$\iff {\sin }^2\left(\alpha \right)+{\left(-\frac{1}{\sqrt{3}}\right)}^2=1$

$\iff {\sin }^2\left(\alpha \right)+\frac{1}{3}=1$

$\iff {\sin }^2\left(\alpha \right)=\frac{2}{3}$

$\iff \sin \left(\alpha \right)=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}$.

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

$=\cos \alpha \cos \frac{\pi }{6}+\sin \alpha \sin \frac{\pi }{6}$

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

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