sinaθ=(m-3)/(m+5),cosa=(4-2m)/(m+5),π/2<θ<π,则m=
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sinaθ=(m-3)/(m+5),cosa=(4-2m)/(m+5),π/2<θ<π,则m=sinaθ=(m-3)/(m+5),cosa=(4-2m)/(m+5),π/2<θ<π,则m=sinaθ=
sinaθ=(m-3)/(m+5),cosa=(4-2m)/(m+5),π/2<θ<π,则m=
sinaθ=(m-3)/(m+5),cosa=(4-2m)/(m+5),π/2<θ<π,则m=
sinaθ=(m-3)/(m+5),cosa=(4-2m)/(m+5),π/2<θ<π,则m=
π/2<θ<π,
sinA>0,cosA0
m3
(4-2m)/(m+5)
sinaθ=(m-3)/(m+5),cosa=(4-2m)/(m+5),π/2<θ<π,则m=
(1)已知sina=m(lml
sinA=(m-3)/(m+5)cosA=(4-2m)/(m+5)A∈(π/2,π)则tanA=
cosa=(4-2m)/(m+5),sina=(m-3)/(m+5)(π/2
已知sina=m-3/m+5,cosa=4-2m/m+5(180度
已知sina=m(|m|
已知sina=(2m-5)/(m+1),cosa=-m/(m+1),且a为第二象限角,则m值为
sina=m(0
已知sinA=((m-3)/(m+5)),cosA=((4-2m)/(m+5)),且(π/2)≤A≤π,则m等于多少
sina=(4-2m)/(m+5),cosa=(m-3)/(m+5),a为4项限角,求tana
已知1/sina的绝对值=-1/sina,且lg(cosa)有意义1. 试判断角a所在的象限2. 若角a的终边与单位圆相交于M(3/5,m),求m的值以及sina的值(各位拜托拜托了~~~~~~急需!)题目的1/sina的绝对值只针对分母~
计算:(m+3m+5m+…+2015m)-(2m+4m+6m+…+2014m)=
若(sina)^2+2*(sinb)^2=2cosa.y=(sina)^2+(sinb)^2的最大值为M,最小值为m,则m+M=
sina-cosa=m,求cosa+sina
已知sina=m-3/m+5,cosa=4-2m/m+5(m≠0),则m=___,tana=____
已知m=(根号5sinA+1)/(cosA+2),求m的取值范围
已知(1/2)sina+(根号3/2)cosa=lg(m-1),求m的取值范围
若∠A为锐角,sinA-(1/3)=2m,则m的取值范围是