Cho góc $a$ thỏa mãn $\pi <a<\frac{3\pi }{2}$ và $\tan a=5$.
1. Tính $\sin a;\cos a;\sin \left(x+\frac{5\pi }{6}\right)$.
2. Tính $\sin 2a;\cos 2a$.
3. Tính $\sin \frac{a}{2};\cos \frac{a}{2}$.

(AI giải đề cho tôi)
Vì $\pi <a<\frac{3\pi }{2}$ nên $a$ ở góc phần tư III: $\sin a<0,\cos a<0$, và $\tan a=5$.

Ta có $\frac{1}{{\cos }^{2}a}=1+{\tan }^{2}a=26\Rightarrow {\cos }^{2}a=\frac{1}{26}$, suy ra $\cos a=-\frac{1}{\sqrt{26}}=-\frac{\sqrt{26}}{26}$; $\sin a=\tan a\cdot \cos a=-\frac{5}{\sqrt{26}}=-\frac{5\sqrt{26}}{26}$.

1) $\sin \left(a+\frac{5\pi }{6}\right)=\sin a\cos \frac{5\pi }{6}+\cos a\sin \frac{5\pi }{6}=\left(-\frac{5}{\sqrt{26}}\right)\left(-\frac{\sqrt{3}}{2}\right)+\left(-\frac{1}{\sqrt{26}}\right)\cdot \frac{1}{2}=\frac{5\sqrt{3}-1}{2\sqrt{26}}=\frac{\sqrt{26}(5\sqrt{3}-1)}{52}$.

2) $\sin 2a=2\sin a\cos a=2\cdot \left(-\frac{5}{\sqrt{26}}\right)\left(-\frac{1}{\sqrt{26}}\right)=\frac{10}{26}=\frac{5}{13}$.
$\cos 2a={\cos }^{2}a-{\sin }^{2}a=\frac{1}{26}-\frac{25}{26}=-\frac{12}{13}$.

3) Vì $\pi <a<\frac{3\pi }{2}$ nên $\frac{\pi }{2}<\frac{a}{2}<\frac{3\pi }{4}$: $\sin \frac{a}{2}>0,\cos \frac{a}{2}<0$.
$\sin \frac{a}{2}=\sqrt{\frac{1-\cos a}{2}}=\sqrt{\frac{1+\frac{1}{\sqrt{26}}}{2}}=\sqrt{\frac{\sqrt{26}+1}{2\sqrt{26}}}=\frac{\sqrt{26+\sqrt{26}}}{2\sqrt{13}}$.
$\cos \frac{a}{2}=-\sqrt{\frac{1+\cos a}{2}}=-\sqrt{\frac{1-\frac{1}{\sqrt{26}}}{2}}=-\sqrt{\frac{\sqrt{26}-1}{2\sqrt{26}}}=-\frac{\sqrt{26-\sqrt{26}}}{2\sqrt{13}}$.

Kiểm tra: ${\sin }^2\frac{a}{2}+{\cos }^2\frac{a}{2}=\frac{\sqrt{26}+1+\sqrt{26}-1}{2\sqrt{26}}=1$ đúng; $2\sin \frac{a}{2}\cos \frac{a}{2}=-\frac{2\sqrt{(26+\sqrt{26})(26-\sqrt{26})}}{52}=-\frac{2\sqrt{676-26}}{52}=-\frac{2\sqrt{650}}{52}=-\frac{10\sqrt{26}}{52}=-\frac{5\sqrt{26}}{26}=\sin a$ đúng.