TAN(a+π/4)=1/2 -π/2<A<0 求(2SINA^2+SIN2A)/COS(A-π/4) 的值,
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/25 11:16:05
TAN(a+π/4)=1/2 -π/2<A<0 求(2SINA^2+SIN2A)/COS(A-π/4) 的值,
tan(A+π/4) =1/2
(tanA+1)/(1-tanA)=1/2
2tanA+2 =1-tanA
tanA= -1/3
sinA = -1/√10
cosA = 3/√10
sin2A = 2sinAcosA = -3/5
cos(A-π/4) = cosAcos(π/4)+sinAsin(π/4)
=(1/√2)(2/√10)
=1/√5
[2(sinA)^2+sin2A]/ cos(A-π/4)
=[2(1/10)-3/5]/(1/√5)
= -2√5/5
(tanA+1)/(1-tanA)=1/2
2tanA+2 =1-tanA
tanA= -1/3
sinA = -1/√10
cosA = 3/√10
sin2A = 2sinAcosA = -3/5
cos(A-π/4) = cosAcos(π/4)+sinAsin(π/4)
=(1/√2)(2/√10)
=1/√5
[2(sinA)^2+sin2A]/ cos(A-π/4)
=[2(1/10)-3/5]/(1/√5)
= -2√5/5
已知tan(π/4+已知tan (π/4+a)=3 求 sin2a-2cos^2a-1的值
已知tan (π/4+a)=3 求 sin2a-2cos^2a的值
tan(π/4+a)=1/2 求(sin2a-cos平方a)÷(1+cos2a)
已知tan(π/4+a)=2,求tana的值,求sin2a+sina平方+cos2a的值
已知tan(π/4+a)=-1/2,求[sin2a-2(cosa)^2]/1+tan a
已知tan(a+4/π)=1/2 ,且-π/2<a<0,则(2sin^2a+sin2a)/cos(a-4/π)=
已知tan(π/4+A)=1/2 ①求tanA的值 ②求(sin2A-cos^A)/(1+cos2A)的值
已知tan(π/4-a)=1/3,a属于(0,π/4).(1)求(a)=(sin2a-2cos^2a)/(1+tana)
1-sin2a-cos^2(a-π/4)
已知tana=2求(1)tan(a+π/4)的值(2)sin2a+cos^2(π-a)除以(1+cos2a)的值
已知sina=-3/5,a∈(π,3/2π)求sin2a/cos²a的值
已知cos2a+sin2a(2sina-1)=2/5,a∈(π/2,π),则tan(a+π/4)的值为