Không dùng máy tính cầm tay, tính các giá trị lượng giác của các góc :

a) $\frac{5\pi }{12}$;

a) Ta có:

+) $\text{cos}\frac{5\pi }{12}=\text{cos}\left(\frac{\pi }{4}+\frac{\pi }{6}\right)=\text{cos}\frac{\pi }{4}.\text{cos}\frac{\pi }{6}-\text{sin}\frac{\pi }{4}.\sin \frac{\pi }{6}=\frac{\sqrt{2}}{2}.\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}.\frac{1}{2}$$=\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}=\frac{\sqrt{6}-\sqrt{2}}{4}$+.

+) +$\sin \frac{5\pi }{12}=\sin \left(\frac{\pi }{4}+\frac{\pi }{6}\right)=\sin \frac{\pi }{4}.\text{cos}\frac{\pi }{6}\text{+\hspace{0.17em}cos}\frac{\pi }{4}.\sin \frac{\pi }{6}=\frac{\sqrt{2}}{2}.\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}.\frac{1}{2}$

$=\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{6}+\sqrt{2}}{4}$.

+) $\tan \frac{5\pi }{12}=\frac{\sin \frac{5\pi }{12}}{\text{cos}\frac{5\pi }{12}}=\frac{\frac{\sqrt{6}-\sqrt{2}}{4}}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}=2-\sqrt{3}$.

+) $\cot \frac{5\pi }{12}=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}$.