2sin ^2(19π/4)+tan^2(8π/3)*tan(-7π/4)
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/23 03:21:14
2sin^2(19π/4)+tan^2(8π/3)*tan(-7π/4)2sin^2(19π/4)+tan^2(8π/3)*tan(-7π/4)2sin^2(19π/4)+tan^2(8π/3)*ta
2sin ^2(19π/4)+tan^2(8π/3)*tan(-7π/4)
2sin ^2(19π/4)+tan^2(8π/3)*tan(-7π/4)
2sin ^2(19π/4)+tan^2(8π/3)*tan(-7π/4)
原式=2sin²(3π/4)+tan²(2π/3)*tan(π/4)
=2(√2/2)²+(-√3)²*1
=4
sin(2a+π/4)化简成tan
2sin ^2(19π/4)+tan^2(8π/3)*tan(-7π/4)
sin(-19π/6)=?已知[3Sin(π+a)+Cos(π-a)]/4Sin(-a)-Sin(5π/2+a)=2,求tan a
若a属于(0,π/2)试比较tanα、tan(tanα)、tan(sinα)
证明tan^θ-sin^2θ=tan^2θsin^2θ证明(1)tan^θ-sin^2θ=tan^2θsin^2θ(2)sin^4x+cos^4x=1-2sin^2cos^2x(31-tan^2x/1+tan^2x=cos^2x-sin^2x
tan(2π/13)+4sin(6π/13)
1)tan(x/2+π/4)+tan(x/2-π/4)=2tanx(2)(1-2sinαcosα/cos²α-sin²α)=(1-tanα)/(1+tanα)
Sinπ/3Tanπ/3+tanπ/6COSπ/6-Tanπ/4COSπ/2
化简sin(2π-α)tan(α+π)tan(-α-π)/
tan(a+π/4)=2,则cos2a+3sin^2a+tan a=
求证2(cosθ -sinθ )/1+sinθ +cosθ =tan(π/4-θ /2)-tanθ /2
已知tanα-tanβ=2tanα^2tanβ,且αβ均不等于kπ/2.试求sinβsin(2α+β)/sinβ或者能给思路已知就是这个:tanα-tanβ=2tanαtanαtanβ
化简sin(9π+α)tan(3/2π+α)/cos(-4π-α)tan(15π+α)tan(7/2π-α)
已知5sinβ=sin(2α+β),求tan(α+β)/tanα
已知tan(α+π/4)=1/3已求出tanα=-1/2求2sin²α-sin(π-α)sin[(π/2)-α]+sin²[(3π/2)+α]的值.
已知sin(a+b)=2/3,sin(a-b)=3/4,求[tan(a+b)-tana-tanb]/tan²b*tan(a+b)的值
求证:(1-sinα+cosα)/(1+sinα+cosα)=tan(π/4-α/2)
求证:1-2sinαcosα/cos²α-sin²α=tan(π/4-α)