Cho \(\sin a = \frac{4}{5},\,\,\cos b = \frac{8}{{17}}\) với \(\frac{\pi }{2}...

* Hướng dẫn giải

Ta có: \(\frac{\pi }{2} < a < \pi  \Rightarrow \cos a < 0\)

\( \Rightarrow \cos a =  - \sqrt {1 - {{\sin }^2}a} \)\( =  - \sqrt {1 - \frac{{16}}{{25}}}  =  - \sqrt {\frac{9}{{25}}}  =  - \frac{3}{5}\)

Ta có: \(0 < b < \frac{\pi }{2} \Rightarrow \sin b > 0\)

\(\begin{array}{l} \Rightarrow \sin b = \sqrt {1 - {{\cos }^2}b} \\ = \sqrt {1 - \frac{{64}}{{289}}}  = \sqrt {\frac{{225}}{{289}}}  = \frac{{15}}{{17}}\\ \Rightarrow \sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\\ = \frac{4}{5}.\frac{8}{{17}} - \frac{3}{5}.\frac{{15}}{{17}} =  - \frac{{13}}{{85}}.\end{array}\)