Đáp án:

Giải thích các bước giải:

1.A=$2\cos  \frac{160^{\circ}+20^{\circ}}{2}.\cos  \frac{160^{\circ}-20^{\circ}}{2}+\sin 30^{\circ}=-2\cos  90^{\circ}.\cos 70^{\circ}+\sin 30^{\circ}=\sin 30^{\circ}=\frac{1}{2}$

2.B=$(-\sin 40^{\circ}+\sin 40^{\circ})+(\sin ^{2}40^{\circ}+\cos ^{2}40^{\circ})=0+1=1$

3.C=$\frac{1+\cos 100^{\circ}}{2}+\frac{1+\cos 80^{\circ}}{2}-\frac{\sin 10^{\circ}.\sin 80^{\circ}}{\cos 10^{\circ}.\cos 80^{\circ}}$

=$\frac{1+\cos 100^{\circ}}{2}+\frac{1+\cos 80^{\circ}}{2}-\frac{\frac{\cos 70^{\circ}}{2}}{\frac{\cos 70^{\circ}}{2}}$

=$1+\frac{1}{2}.(\cos 100^{\circ}+\cos 80^{\circ})-1$

=$\frac{1}{2}.2.\cos 90^{\circ}.\cos 10^{\circ}=0$

1. 

$A=\cos 160^o+\cos 20^o+\sin 30^o$

$=2\cos 90^o.\cos 70^o+\sin 30^o$

$=\sin 30^o$

2.

$B=\sin(-40^o)+\sin 40^o + \sin^240^o+\cos^240^o$

$= 0+1=1$

3.

$C=\cos^250^o+\cos^240^o-\tan 10^o.\tan 80^o$

$= \cos^250^o+\sin^250^o-\tan 10^o.\cot 10^o$

$= 1-1=0$