化简[ sec(-θ)+cos(-θ+180° ) ] / [ cos(180 °-θ)+sin(360 °-θ ) ]
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/19 11:28:55
化简[ sec(-θ)+cos(-θ+180° ) ] / [ cos(180 °-θ)+sin(360 °-θ ) ]
[ sec(-θ)+cos(-θ+180° ) ] / [ cos(180 °-θ)+sin(360 °-θ ) ]
=(1/cosθ-cosθ)/(cosθ-sinθ)
=1+sinθ/cosθ-(cosθ)^2+sinθcosθ
=tanθ+(sinθ)^2+sinθcosθ
=tanθ+1/2(1-cos2θ)+1/2sin2θ
=tanθ+1/2+√2/2(sin2θcos45°-cos2θsin45°)
=tanθ+√2/2sin(2θ-45°)+1/2
=(1/cosθ-cosθ)/(cosθ-sinθ)
=1+sinθ/cosθ-(cosθ)^2+sinθcosθ
=tanθ+(sinθ)^2+sinθcosθ
=tanθ+1/2(1-cos2θ)+1/2sin2θ
=tanθ+1/2+√2/2(sin2θcos45°-cos2θsin45°)
=tanθ+√2/2sin(2θ-45°)+1/2
化简[ sec(-θ)+cos(-θ+180° ) ] / [ cos(180 °-θ)+sin(360 °-θ ) ]
2cos³θ+sin²(360°-θ)+cos(360°-θ)-3/2+2cos²(180°-θ)+cos(-θ)怎样化
[2cos³θ+sin²(360°-θ)+cos(360°-θ)-3]/[2+2cos²(180°-θ)+cos(-θ)
化简:cos平方(θ+15°)+sin平方(θ-15°)+sin(θ+180°)cos(θ-180°)
化简sinθ-cosθ除以(1-tanθ)可得 A -sinθ B -cosθ C -secθ D cscθ
化简:cos(90°+α)·csc(370°+α)·tan(180°-α)/sec(360°-α)·sin(180°+α
化简:1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ
化简(sinθcosθ+cos平方θ)/(sin平方θ+cos平方θ)
sin(θ+75°)+cos(θ+45°)-3cos(θ+15°
Cos(180°+α)*sin(α+360°)/sin(-α-180°)*cos(-180°-α)
[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]=?
sinΘ-2cosΘ=0,Θ属于(0.,90°),求sinΘ、cosΘ的值.