Đáp án:$a)sin(\pi-\alpha )=\dfrac{2}{3}\\
b) sin(\alpha-\pi)=-\dfrac{2}{3}\\
c) cos(\pi+\alpha)=-\dfrac{\sqrt{5}}{3}\\
d) tan(5\pi+\alpha)=\dfrac{2\sqrt{5}}{5}$
Giải thích các bước giải:
$sin^2\alpha+cos^2\alpha=1\\
\Rightarrow cos^2\alpha=1-sin^2\alpha=1-\dfrac{2}{3}^2=\dfrac{5}{9}\\
cosx\alpha=\pm \dfrac{\sqrt{5}}{3}$
Do $0<\alpha<\dfrac{\pi}{2}$
Nên $cos\alpha=\dfrac{\sqrt{5}}{3}$
$a)sin(\pi-\alpha )=sin\alpha =\dfrac{2}{3}\\
b) sin(\alpha-\pi)=sin\left [ -(\pi-\alpha) \right ]=-sin\alpha=-\dfrac{2}{3}\\
c) cos(\pi+\alpha)=-cos\alpha=-\dfrac{\sqrt{5}}{3}\\
d) tan(5\pi+\alpha)=tan(\pi+4\pi+\alpha)=tan(4\pi+\alpha)=tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{2}{3}}{\dfrac{\sqrt{5}}{3}}=\dfrac{2\sqrt{5}}{5}$
