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

b) – 555°.

b) Ta có:

– 555° = $\frac{\pi .\left(-555°\right)}{180°}=-\frac{37\pi }{12}=-\left(3\pi +\frac{\pi }{12}\right)$ rad.

Khi đó:

+) $\text{cos}\left(-555°\right)=\text{cos}\left(3\pi +\frac{\pi }{12}\right)=-\text{cos}\left(\frac{\pi }{12}\right)=-\text{cos}\left(\frac{\pi }{3}-\frac{\pi }{4}\right)$

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

+) $\sin \left(-555°\right)=\sin \left(3\pi +\frac{\pi }{12}\right)=\sin \left(\frac{\pi }{12}\right)=\sin \left(\frac{\pi }{3}-\frac{\pi }{4}\right)$

$=\sin \frac{\pi }{3}\text{cos}\frac{\pi }{4}-\text{cos}\frac{\pi }{3}\sin \frac{\pi }{4}=\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}-\frac{1}{2}.\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$.

+) $\tan \left(-555°\right)=\frac{\sin \left(-555°\right)}{\text{cos}\left(-555°\right)}=\frac{\frac{\sqrt{6}-\sqrt{2}}{4}}{-\frac{\sqrt{6}+\sqrt{2}}{4}}=-2+\sqrt{3}$.

+)  $\cot \left(-555°\right)=\frac{1}{-2+\sqrt{3}}=-2-\sqrt{3}$.