Đáp án:
\(\begin{array}{l}
\sin x = \dfrac{4}{5}\\
\sin \left( {x + \dfrac{\pi }{3}} \right) = \dfrac{{4 - 3\sqrt 3 }}{{10}}\\
\cos \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{{ - 7\sqrt 2 }}{{10}}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{\pi }{2} < x < \pi \Rightarrow \sin x > 0\\
{\sin ^2}x + {\cos ^2}x = 1\\
\Leftrightarrow {\sin ^2}x + {\left( { - \dfrac{3}{5}} \right)^2} = 1\\
\Leftrightarrow {\sin ^2}x + \dfrac{9}{{25}} = 1\\
\Leftrightarrow {\sin ^2}x = \dfrac{{16}}{{25}}\\
\sin x > 0 \Rightarrow \sin x = \dfrac{4}{5}\\
\sin \left( {x + \dfrac{\pi }{3}} \right) = \sin x.\cos \dfrac{\pi }{3} + \cos x.\sin \dfrac{\pi }{3}\\
= \dfrac{4}{5}.\dfrac{1}{2} + \dfrac{{ - 3}}{5}.\dfrac{{\sqrt 3 }}{2} = \dfrac{{4 - 3\sqrt 3 }}{{10}}\\
\cos \left( {x + \dfrac{\pi }{4}} \right) = \cos x.\cos \dfrac{\pi }{4} - \sin x.\sin \dfrac{\pi }{4}\\
= \dfrac{{ - 3}}{5}.\dfrac{{\sqrt 2 }}{2} - \dfrac{4}{5}.\dfrac{{\sqrt 2 }}{2} = \dfrac{{ - 7\sqrt 2 }}{{10}}
\end{array}\)
