12 câu Trắc nghiệm Giá trị lượng giác của một góc bất kì từ 0o đến 150o có đáp án (Nhận biết)

$\sin ({180}^{\circ }-\alpha )=- \cos \alpha$

$\sin ({180}^{\circ }-\alpha )=-\sin \alpha$

$\sin ({180}^{\circ }-\alpha )=\sin \alpha$

$\sin ({180}^{\circ }-\alpha )=\cos \alpha$

sin α < 0

cos α > 0

tan α > 0

cot α < 0

sin α = cos β

tan α = cot β

$cot\beta =\frac{1}{\cot \alpha }$

cos α = − sin β

sin α = − cos β

cos α = sin β

tan α = cot β

cot α = tan β

$\sin {90}^{\circ } < \sin {100}^{\circ }$

$\cos {95}^{\circ } > \cos {100}^{\circ }$

$\tan {85}^{\circ } < \tan {125}^{\circ }$

$\cos {145}^{\circ } > \cos {125}^{\circ }$

$\sin {90}^{\circ } < \sin {150}^{\circ }$

$\sin {90}^{\circ }15\prime < \sin {90}^{\circ }30\prime$

$\cos {90}^{\circ }30\prime > \cos {100}^{\circ }$

$\cos {150}^{\circ } > \cos {120}^{\circ }$

2

0

$\sqrt{3}$

1

1

$\sqrt{2}$

$\sqrt{3}$

0

${\sin }^2\alpha +\cos {\alpha }^2=1$

${\sin }^2\alpha +{\cos }^2\frac{\alpha }{2}=1$

$\sin {\alpha }^2+\cos {\alpha }^2=1$

${\sin }^{2}2\alpha +{\cos }^{2}2\alpha =1$

${\sin }^2\alpha +{\cos }^2\alpha =1$

$1+{\cot }^2\alpha =\frac{1}{{\sin }^2\alpha }\left(\sin \alpha \ne 0\right)$

$\tan \alpha .\cot \alpha =-1\left(\sin \alpha .\cos \alpha \ne 0\right)$

$1+{\tan }^2\alpha =\frac{1}{{\cos }^2\alpha }\left(\cos \alpha \ne 0\right)$

${\cos }^2\frac{\alpha }{2}+{\sin }^2\frac{\alpha }{2}=\frac{1}{2}$

${\cos }^2\frac{\alpha }{3}+{\sin }^2\frac{\alpha }{3}=\frac{1}{3}$

${\cos }^2\frac{\alpha }{4}+{\sin }^2\frac{\alpha }{4}=\frac{1}{4}$

$5\left({\cos }^2\frac{\alpha }{5}+{\sin }^2\frac{\alpha }{5}\right)=5$