Question

高中数学 [sin(-250°)cos70°]/[(cos155°)^2-(sin25°)^2]的值为？拜托了！

Answer 1

【参考答案】

原式=[sin(360°-250°)cos70°]/[(-cos25°)²-(sin25°)²]
=[sin110°cos70°]/(cos25°)²-(sin25°)²]
=[sin70°cos70°]/cos50°
=(1/2)sin140°/cos50°
=(1/2)sin40°/cos50°
=(1/2)sin40°/sin40°
=1/2

Answer 2

[sin(-250°)cos70°]/[(cos155°)^2-(sin25°)^2]
=（sin110°cos70°）/{(cos155°)^2-[sin(π-155°)]^2}
=sin(π-70°)cos70°/[(cos155°)^2-(-sin155°)^2]
=[sin70°cos70°]/[cos(310°)]
=[sin70°cos70°]/[cos70°]
=sin70°

Answer 3

sin(-250°)=sin(-70°-π)=-sin(-70°)=-sin70°
cos155°=cos(π-25°)=-cos(-25°)=-cos25°
[sin(-250°)cos70°]=sin70°cos70°=1/2sin140°
[(cos155°)^2-(sin25°)^2]=(cos25°)^2-(sin25°)^2=cos50°
sin140°=sin(50°+π/2)=cos50°
所以1/2sin140°=1/2cos50°
原式=1/2cos50° ÷ cos50°=1/2

Answer 4

具体过程 原式化为：[sin(70°）cos(70°)]/cos(25°+25°)
=sin(140°)/2cos50°
=sin40°/2cos50°
=1/2

Answer 5

=sin70°cos70°/[(cos25°)^2-(sin25°)^2]
=sin140°/2cos50°
=sin40°/2cos50°
=1/2

Answer 6

1/2 (希望没算错）