Đáp án:
1a)
$A=0\\
b)
\sin2a=\dfrac{-4\sqrt{2}}{9}\\
\cos2a=\dfrac{7}{9}$
Giải thích các bước giải:
$1a)
A=\dfrac{4\cos^2\alpha-\sin^2\alpha}{5\cos^2\alpha+4}\\
=\dfrac{4\dfrac{\cos^2\alpha}{\cos^2\alpha}-\dfrac{\sin^2\alpha}{\cos^2\alpha}}{\dfrac{5\cos^2\alpha}{\cos^2\alpha}+\dfrac{4}{\cos^2\alpha}}$
$=\dfrac{4-\tan^2\alpha}{5+4(1+\tan^2\alpha)}$(1)
Thay $\tan\alpha=2$ vào (1) ta được
$A=\dfrac{4-2^2}{5+4(1+2^2)}=\dfrac{0}{25}=0\\
b)
\sin^2a+\cos^2a=1\\
\Rightarrow \cos^2a=1-\sin^2a\\
=1-\left (\dfrac{1}{3} \right )^2=\dfrac{8}{9}\\
\Rightarrow \cos a=\pm \dfrac{2\sqrt{2}}{3}$
Do $\dfrac{\pi}{2}<a<\pi\Rightarrow \cos a<0$
$\Rightarrow \cos a=\dfrac{-2\sqrt{2}}{3}\\
\Rightarrow \sin2a=2\sin a\cos a=2.\dfrac{1}{3}.\dfrac{-2\sqrt{2}}{3}=\dfrac{-4\sqrt{2}}{9}\\
\Rightarrow \cos2a=1-2\sin^2a=1-2.\left (\dfrac{1}{3} \right )^2=\dfrac{7}{9}$
