Tính các giá trị lượng giác của góc 2α, biết:

b)  $\sin \frac{\alpha }{2}=\frac{3}{4}$ và  $\pi <\alpha <2\pi$.

b) Ta có:  $\text{cos}\frac{\alpha }{2}=-\sqrt{1-{\left(\frac{3}{4}\right)}^2}=-\frac{\sqrt{7}}{4}$ (vì  $\pi <\alpha <2\pi \iff \frac{\pi }{2}<\frac{\alpha }{2}<\pi$).

Khi đó:

 $\sin \alpha =2.\sin \frac{\alpha }{2}.\text{cos}\frac{\alpha }{2}=2.\frac{3}{4}.\left(-\frac{\sqrt{7}}{4}\right)=-\frac{3\sqrt{7}}{8}$;

 $\text{cos}\alpha =2.{\text{cos}}^2\frac{\alpha }{2}-1=2.{\left(-\frac{\sqrt{7}}{4}\right)}^2-1=-\frac{1}{8}$;

 $\sin 2\alpha =2.\sin \alpha .\text{cos}\alpha =2.\left(-\frac{3\sqrt{7}}{8}\right).\left(-\frac{1}{8}\right)=\frac{3\sqrt{7}}{32}$;

 $\text{cos}2\alpha =2.{\text{cos}}^2\alpha -1=2.{\left(-\frac{1}{8}\right)}^2-1=-\frac{31}{32}$;

 $\tan 2\alpha =\frac{\sin 2\alpha }{\text{cos\hspace{0.17em}}2\alpha }=\frac{\frac{3\sqrt{7}}{8}}{-\frac{31}{32}}=-\frac{12\sqrt{7}}{31}$;

 $\cot 2\alpha =\frac{1}{\tan 2\alpha }=-\frac{31}{12\sqrt{7}}$.