Biết \(\sin a = \dfrac{5}{{13}};\cos b = \dfrac{3}{5}\left( {\dfrac{\pi }{2} < a < \pi ;0 < b <
Biết \(\sin a = \dfrac{5}{{13}};\cos b = \dfrac{3}{5}\left( {\dfrac{\pi }{2} < a < \pi ;0 < b < \dfrac{\pi }{2}} \right).\) Hãy tính \(\sin \left( {a + b} \right)\)
Đáp án đúng là: C
\({\sin ^2}a + {\cos ^2}a = 1\)
Ta có \(\left\{ \begin{array}{l}{\sin ^2}a + {\cos ^2}a = 1\\{\sin ^2}b + {\cos ^2}b = 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\dfrac{{25}}{{169}} + {\cos ^2}a = 1\\{\sin ^2}b + \dfrac{9}{{25}} = 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\cos ^2}a = \dfrac{{144}}{{169}}\\{\sin ^2}b = \dfrac{{16}}{{25}}\end{array} \right.\)
Do \(\dfrac{\pi }{2} < a < \pi ;0 < b < \dfrac{\pi }{2}\)\( \Rightarrow \cos a < 0;\sin b > 0\)
\( \Rightarrow \cos a = - \dfrac{{12}}{{13}};\sin b = \dfrac{4}{5}\)\( \Rightarrow \sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b = \dfrac{5}{{13}}.\dfrac{3}{5} - \dfrac{{12}}{{13}}.\dfrac{4}{5} = - \dfrac{{33}}{{65}}\).
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