Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\pi < x < \dfrac{{3\pi }}{2} \Leftrightarrow \left\{ \begin{array}{l}
\sin x < 0\\
\cos x < 0
\end{array} \right.\\
{\sin ^2}x + {\cos ^2}x = 1\\
\cos x < 0 \Rightarrow \cos x = - \sqrt {1 - {{\sin }^2}x} = - \dfrac{{\sqrt 5 }}{3}\\
\cos \left( {x - \dfrac{\pi }{6}} \right) = \cos x.\cos \dfrac{\pi }{6} + \sin x.\sin \dfrac{\pi }{6}\\
= \left( { - \dfrac{{\sqrt 5 }}{3}} \right).\dfrac{{\sqrt 3 }}{2} + \dfrac{{ - 2}}{3}.\dfrac{1}{2} = \dfrac{{ - 2 - \sqrt {15} }}{6}\\
b,\\
A = 4\left( {{{\sin }^4}\alpha + {{\cos }^4}\alpha } \right) - \cos 4\alpha \\
= 4.\left[ {{{\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)}^2} - 2{{\sin }^2}\alpha .{{\cos }^2}\alpha } \right] - \left( {2{{\cos }^2}2\alpha - 1} \right)\\
= 4.\left( {{1^2} - \dfrac{1}{2}.{{\left( {2\sin \alpha .\cos \alpha } \right)}^2}} \right) - \left( {2{{\cos }^2}2\alpha - 1} \right)\\
= 4 - 2.{\sin ^2}2\alpha - 2{\cos ^2}2\alpha - 1\\
= 3 - 2.\left( {{{\sin }^2}2\alpha + {{\cos }^2}2\alpha } \right)\\
= 3 - 2.1 = 1\\
c,\\
\dfrac{{\sin 2x - 2\sin x}}{{\sin 2x + 2\sin x}} = \dfrac{{2\sin x.\cos x - 2\sin x}}{{2\sin x.\cos x + 2\sin x}}\\
= \dfrac{{2\sin x\left( {\cos x - 1} \right)}}{{2\sin x.\left( {\cos x + 1} \right)}} = \dfrac{{\cos x - 1}}{{\cos x + 1}}\\
= \dfrac{{\left( {1 - 2{{\sin }^2}\dfrac{x}{2}} \right) - 1}}{{\left( {2{{\cos }^2}\dfrac{x}{2} - 1} \right) + 1}} = \dfrac{{ - {{\sin }^2}\dfrac{x}{2}}}{{{{\cos }^2}\dfrac{x}{2}}} = - {\tan ^2}\dfrac{x}{2}
\end{array}\)
