Rút gọn biểu thức ${\rm{S}} = \cos {\rm{(9}}{{\rm{0}}^0} - x)\sin \left( {{{180}^0} - x} \right) $ $- {\rm{\sin (9}}{{\rm{0}}^0} - x)\cos \left( {{{180}^0} - x} \right)$ ta được kết quả:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$0$     

$1$     

$2$                 

$4$

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

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

$\tan {225^0} =  - 1$

$\cot {225^0} = 1$