Cho \(\cos \alpha  = \dfrac{1}{3}\). Khi đó \(\cos \left( {3\pi  + \alpha } \right)\) bằng:

$\cot (\pi  - \alpha ) = \cot (\pi  + \alpha )$     

$\cot (\dfrac{\pi }{2} - \alpha ) = \tan (\pi  + \alpha )$

$\cot (\pi  + \alpha ) = \cot ( - \alpha )$

$\cot (\pi  + \alpha ) = \cot (\dfrac{\pi }{2} - \alpha )$

$\sin (\pi  - \alpha ) = \sin (\pi  + \alpha )$

$\sin (\dfrac{\pi }{2} - \alpha ) = \sin (\pi  + \alpha )$

$\sin (\pi  + \alpha ) = \sin ( - \alpha )$

$\sin (\pi  + \alpha ) = \cos (\dfrac{\pi }{2} - \alpha )$                                  

$\tan (\pi  - \alpha ) = \tan (\dfrac{\pi }{2} - \alpha )$

$\tan (\dfrac{\pi }{2} - \alpha ) = \tan ( - \alpha )$

$\tan (\pi  - \alpha ) = \tan (\dfrac{\pi }{2} + \alpha )$

$\tan (\dfrac{\pi }{2} - \alpha ) = \cot (\pi  + \alpha )$

$\tan (\pi  + \alpha ) =  - \tan (\alpha )$

$\tan ( - \alpha ) =  - \tan (\alpha )$

$\tan (\dfrac{\pi }{2} - \alpha ) = \cot (\alpha )$

$\tan (\pi  - \alpha ) =  - \tan (\alpha )$

$0$     

$1$     

$2$                 

$4$

$S = 1$

$S = 0$

$S = {\sin^2}x-{\cos^2}x$    

$S = 2\sin x\cos x$

$\sin {225^0} = \dfrac{{ - \sqrt 2 }}{2}$

${\rm{cos22}}{{\rm{5}}^0} =  - \dfrac{{\sqrt 2 }}{2}$

$\tan {225^0} =  - 1$

$\cot {225^0} = 1$