Question

Cho \(\sin \alpha  = \frac{{12}}{{13}}\) và \(\cos \alpha  =  - \frac{5}{{13}}\). Tính \(\sin \left( { - \frac{{15\pi }}{2} - \alpha } \right) - \cos \left( {13\pi  + \alpha } \right)\)

Answer 1

Ta có:

 \(\begin{array}{l}\sin \left( { - \frac{{15\pi }}{2} - \alpha } \right) - \cos \left( {13\pi  + \alpha } \right) =  \sin \left( { -\frac{{16\pi }}{2} +\frac{{\pi }}{2}  + \alpha } \right) - \cos \left( {12\pi  + \pi + \alpha } \right) =  \sin \left( {-8\pi  + \frac{\pi }{2} - \alpha } \right) - \cos \left( { \pi + \alpha } \right) \\ = \sin \left( {\frac{\pi }{2} - \alpha } \right) + \cos \left( \alpha  \right) = \cos \left( \alpha  \right) + \cos \left( \alpha  \right) = 2\cos \left( \alpha  \right) = 2.\left( { - \frac{5}{{13}}} \right) = \frac{{ - 10}}{{13}}\end{array}\)