Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = {\cos ^2}\dfrac{\pi }{{16}} + {\cos ^2}\dfrac{{3\pi }}{{16}} + {\cos ^2}\dfrac{{5\pi }}{{16}} + {\cos ^2}\dfrac{{7\pi }}{{16}}\\
= {\cos ^2}\dfrac{\pi }{{16}} + {\cos ^2}\dfrac{{3\pi }}{{16}} + {\sin ^2}\left( {\dfrac{\pi }{2} - \dfrac{{5\pi }}{{16}}} \right) + {\sin ^2}\left( {\dfrac{\pi }{2} - \dfrac{{7\pi }}{{16}}} \right)\\
= {\cos ^2}\dfrac{\pi }{{16}} + {\cos ^2}\dfrac{{3\pi }}{{16}} + {\sin ^2}\dfrac{{3\pi }}{{16}} + {\sin ^2}\dfrac{\pi }{{16}}\\
= \left( {{{\cos }^2}\dfrac{\pi }{{16}} + {{\sin }^2}\dfrac{\pi }{{16}}} \right) + \left( {{{\cos }^2}\dfrac{{3\pi }}{{16}} + {{\sin }^2}\dfrac{{3\pi }}{{16}}} \right)\\
= 1 + 1 = 2\\
b,\\
\pi < \alpha < \dfrac{{3\pi }}{2} \Rightarrow \left\{ \begin{array}{l}
\sin \alpha < 0\\
\cos \alpha < 0
\end{array} \right.\\
\cos \alpha < 0 \Rightarrow \cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \dfrac{4}{5}\\
B = {\cos ^2}\alpha - 2{\sin ^2}\alpha + \sin 2\alpha \\
= {\cos ^2}\alpha - 2{\sin ^2}\alpha + 2\sin \alpha .\cos \alpha \\
= {\left( { - \dfrac{4}{5}} \right)^2} - 2.{\left( { - \dfrac{3}{5}} \right)^2} + 2.\left( { - \dfrac{4}{5}} \right).\left( { - \dfrac{3}{5}} \right)\\
= \dfrac{{22}}{{25}}\\
c,\\
{\sin ^2}\alpha = 1 - {\cos ^2}\alpha = \dfrac{{16}}{{25}}\\
C = {\cos ^3}\alpha - {\sin ^2}\alpha + 3\cos 2\alpha \\
= {\cos ^3}\alpha - {\sin ^2}\alpha + 3\left( {2{{\cos }^2}\alpha - 1} \right)\\
= {\left( {\dfrac{3}{5}} \right)^3} - \dfrac{{16}}{{25}} + 3.\left( {2.{{\left( {\dfrac{3}{5}} \right)}^2} - 1} \right)\\
= - \dfrac{{158}}{{125}}
\end{array}\)
