Đáp án:
$1)
\sin2\alpha=-\dfrac{\sqrt{3}}{2}\\
\cos2\alpha=-\dfrac{1}{2}\\
2)
+) \cos\alpha=- \dfrac{2\sqrt{2}}{3}\\
+) \tan \alpha=\dfrac{-\sqrt{2}}{4}\\
+)\cot \alpha=-2\sqrt{2}$
Giải thích các bước giải:
$1)
\sin^2\alpha+\cos^2\alpha=1\\
\Rightarrow \sin^2\alpha=1-\cos^2\alpha\\
=1-\dfrac{1}{2}^2=\dfrac{3}{4}\\
\Rightarrow \sin \alpha=\pm \dfrac{\sqrt{3}}{2}$
Do $\dfrac{3\pi}{2}<\alpha<2\pi\Rightarrow \sin \alpha<0$
$\Rightarrow \sin\alpha=-\dfrac{\sqrt{3}}{2}\\
\sin2\alpha=2\sin \alpha\cos \alpha=2.-\dfrac{\sqrt{3}}{2}.\dfrac{1}{2}=-\dfrac{\sqrt{3}}{2}\\
\cos2\alpha=\cos^2\alpha-\sin^2\alpha=\dfrac{1}{2}^2-(\dfrac{-\sqrt{3}}{2})^2=-\dfrac{1}{2}\\
2)
\sin^2\alpha+\cos^2\alpha=1\\
\Rightarrow \cos^2\alpha=1-\sin^2\alpha\\
=1-\dfrac{1}{3}^2=\dfrac{8}{9}\\
\Rightarrow \cos\alpha=\pm \dfrac{2\sqrt{2}}{3}$
Do $\dfrac{\pi}{2}<\alpha<\pi\Rightarrow \cos\alpha<0$
$\Rightarrow \cos\alpha=- \dfrac{2\sqrt{2}}{3}\\
+) \tan \alpha=\dfrac{\sin \alpha}{\cos \alpha}=\dfrac{\dfrac{1}{3}}{- \dfrac{2\sqrt{2}}{3}}=\dfrac{-\sqrt{2}}{4}\\
+)\cot \alpha=\dfrac{1}{\tan\alpha}=-2\sqrt{2}$
3)
$VT=\dfrac{\cos2a-\cos a+1}{\sin2a-\sin a}\\
=\dfrac{\cos^2a-\sin^2a-\cos a+\sin^2a+\cos^2a}{2\sin a.\cos a-\sin a}\\
=\dfrac{\cos a(2\cos a-1)}{\sin a(2\cos a-1)}\\
=\cot a=VT\Rightarrow đpcm$
