cos(α-π)cot(5π-α)/tan(2π-α)sin(-2π-α) 急昂
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/16 00:51:16
x
0EWqy.@A 4Ɗ= Jl2 =3;c!M,ZG.먖IbCDT>?qv@"Z`
+A@Fr5
Ӗ/4pR>mGl{i&GD[.e#~
cos(α-π)cot(5π-α)/tan(2π-α)sin(-2π-α) 急昂
cos(α-π)cot(5π-α)/tan(2π-α)sin(-2π-α) 急昂
cos(α-π)cot(5π-α)/tan(2π-α)sin(-2π-α) 急昂
cos(α-π)cot(5π-α)/tan(2π-α)sin(-2π-α)
=[(-cosa)(-cota)]/[(-tana)(-sina)]
=[cosacota]/[tanasina]=(cota)^3
-cos -cot/-tan -sin
=cos cot/tan sin
=(cos^2/sin)/(sin^2/cos)
=cos^3/sin^3
=cot^3
化简cos(α-π)cot(5π-α)/tan(2π-α)sin(-2π-α)
若COSα=4/5,α∈(0 π)则COtα的值
cosα=4/5,α∈(0,π),则cotα的值等于多少?
求证(cotα-cosα)/cotα×cosα= (cotα×cosα)/ (cotα-cosα)
{[sin(π-α)-cosα]^2}/{cot[(3π/2)+α]} - 2cos(α+5π/6)*cos(α-5π/6)
若sinα-cosα=-1/5,α∈(3/2π,7/4π),求sinα与cosα 若sin(α+β)=1/2,sin(α-β)=1/3,求log√5(ta若sinα-cosα=-1/5,α∈(3/2π,7/4π),求sinα与cosα 若sin(α+β)=1/2,sin(α-β)=1/3,求log√5(tan*cotβ)若sinα-2cosα/3sinα+5
cos(α-π)cot(5π-α)/tan(2π-α)sin(-2π-α) 急昂
化简[sin(π-α)tan(5π/2+α)]/[cos(2π-α)cot(π/2-α)]分后给
已知α∈(0,π),sinα+cosα=1/5,求tanα-cotα的值,
cos(α-3π/2)=1/5,求sinα*cotα α是第三象限角.
sinα+cosα=1/5,α属于(0,π),求cotα的值
sin(π+α)=1/2,求sin(2π-α)-cot(α-π)cosα
cot(π+α)=cotα
已知cot=m,π<α<(3π/2),则cosα=
sin^2α*tanα+cos^2α*cotα-2sinα*cosα(π-α)
化简:(cos^2α-sin^2α)/【2*cot(π/4+α)*cos^2(π/4-α)】
化简[cos(α-π)*cot(5π-α)]/[sin(-2π-α)]②sine*根号下(1+cot²e)(e为第三象限角)
[sinπ(/2+α)-cos(3π/2-α)]/[tan(2kπ -α)+cot(-kπ+α)]=[sin(4kπ-α)sin(π/2 -α)]/[cos(5π+α)-cos(π/2+α)