Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{{ - \pi }}{2} < \alpha < 0 \Rightarrow \left\{ \begin{array}{l}
\sin \alpha < 0\\
\cos \alpha > 0
\end{array} \right.\\
a,\\
\cos \alpha > 0 \Rightarrow \cos \alpha = \sqrt {1 - {{\sin }^2}\alpha } = \dfrac{{4\sqrt 3 }}{7}\\
\cos \left( {\alpha - \dfrac{{5\pi }}{6}} \right) = \cos \alpha .\cos \dfrac{{5\pi }}{6} + \sin \alpha .sin\dfrac{{5\pi }}{6} = \dfrac{{4\sqrt 3 }}{7}.\dfrac{{ - \sqrt 3 }}{2} + \dfrac{{ - 1}}{7}.\dfrac{1}{2} = \dfrac{{ - 13}}{{14}}\\
b,\\
\sin \left( {\dfrac{{2\pi }}{3} - \alpha } \right) = \sin \dfrac{{2\pi }}{3}.\cos \alpha - \sin \alpha .\cos \dfrac{{2\pi }}{3} = \dfrac{{\sqrt 3 }}{2}.\dfrac{{4\sqrt 3 }}{7} - \dfrac{{ - 1}}{7}.\dfrac{{ - 1}}{2} = \dfrac{6}{7}\\
c,\\
\tan \left( {\dfrac{{3\pi }}{4} + \alpha } \right) = \dfrac{{\sin \left( {\dfrac{{3\pi }}{4} + \alpha } \right)}}{{\cos \left( {\dfrac{{3\pi }}{4} + \alpha } \right)}} = \dfrac{{\sin \dfrac{{3\pi }}{4}.\cos \alpha + \cos \dfrac{{3\pi }}{4}.\sin \alpha }}{{\cos \dfrac{{3\pi }}{4}.\cos \alpha - sin\dfrac{{3\pi }}{4}.\sin \alpha }} = \dfrac{{\dfrac{{\sqrt 2 }}{2}.\dfrac{{4\sqrt 3 }}{7} + \dfrac{{ - \sqrt 2 }}{2}.\dfrac{{ - 1}}{7}}}{{\dfrac{{ - \sqrt 2 }}{2}.\dfrac{{4\sqrt 3 }}{7} - \dfrac{{ - 1}}{7}.\dfrac{{\sqrt 2 }}{2}}} = \dfrac{{4\sqrt 3 + 1}}{{1 - 4\sqrt 3 }}
\end{array}\)
