Đáp án:
1) $\cos \dfrac{\alpha }{2} = \dfrac{{\sqrt 5 }}{5};\cos 2\alpha = \dfrac{{ - 7}}{{25}}$
2) $A = 1 + \dfrac{1}{2}\sin 2x$
Giải thích các bước giải:
1) Ta có:
$\sin \alpha = \dfrac{4}{5}$
Do $\dfrac{\pi }{2} < \alpha < \pi \Rightarrow \cos \alpha < 0;\cos \dfrac{\alpha }{2} > 0$
$ \Rightarrow \left\{ \begin{array}{l}
\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \dfrac{3}{5}\\
\cos \dfrac{\alpha }{2} = \sqrt {\dfrac{{1 + \cos \alpha }}{2}} = \dfrac{{\sqrt 5 }}{5}\\
\cos 2\alpha = 2{\cos ^2}\alpha - 1 = \dfrac{{ - 7}}{{25}}
\end{array} \right.$
2) Ta có;
$\begin{array}{l}
A = \dfrac{{{{\cos }^2}x}}{{1 - \tan x}} + \dfrac{{{{\sin }^2}x}}{{1 - \cot x}}\\
= \dfrac{{{{\cos }^2}x}}{{1 - \dfrac{{\sin x}}{{\cos x}}}} + \dfrac{{{{\sin }^2}x}}{{1 - \dfrac{{\cos x}}{{\sin x}}}}\\
= \dfrac{{{{\cos }^3}x}}{{\cos x - \sin x}} + \dfrac{{{{\sin }^3}x}}{{\sin x - \cos x}}\\
= \dfrac{{{{\cos }^3}x - {{\sin }^3}x}}{{\cos x - \sin x}}\\
= \dfrac{{\left( {\cos x - \sin x} \right)\left( {{{\cos }^2}x + \cos x\sin x + {{\sin }^2}x} \right)}}{{\cos x - \sin x}}\\
= {\cos ^2}x + \cos x\sin x + {\sin ^2}x\\
= 1 + \dfrac{1}{2}\sin 2x
\end{array}$
