"已知sin(π 4 α)=2 3"
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/10 09:09:55
cos(α-π/6)+sinα=cosα*√3/2+1/2*sinα+sinα=cosα*√3/2+3/2*sinα=√3(sinπ/6cosα+cosπ/6sinα)=√3sin(π/6+α)=4/
因为:tan(a+π/4)=[tana+tanπ/4]/[1-tanatanπ/4]=(tana+1)/(1-tana)=3,解得:tana=1/2(cosα+2sinα)/(cosα-sinα)=(
(1)由sin(π+α)=√3/2∴sinα=-√3/2,cosα=-1/2∴sin(3π/4-α)=sin3π/4cosα-cos3π/4sinα=(√2/2)×(-1/2)-(-√2/2)(-√3
两端平方得[sin(π/4-a)]^2=25/169,即[1-cos(π/2-2a)]/2=25/169,所以[1-sin(2a)]/2=25/169,则sin(2a)=119/169,由于0再问:√
由题意知sin α-cosα=12,两边平方可得sin2α=34,所以(sin α+cos α)2=1+sin2α=74,又α∈(0,π2),所以sin α+c
sin(π/4+α)*sin(π/4-α)=sin(π/4+α)*cos(π/4-α)=1/2sin(π/2+2α)=1/2cos2α1/2cos2α=1/4cos2α=1/2α∈(π/4,π/2),
⑴(sinα+2cosα)/(5cosα-sinα)=(2cosα+2cosα)/(5cosα-2cosα)=(4cosα)/(3cosα)=4/3⑵tan(α+π/4)=sin(α+π/4)/cos
[sinα+2cos(5π/2+α)]/[cos(π-α)-sin(π/2-α)]=-1/4即(sinα-2sinα)/(-cosα-cosα)=-1/4,所以sinα/(2cosα)=-1/4,所以
∵2sin²α-sinαcosα-3cos²α=0,∴(2sinα-3cosα)(sinα+cosα)=0.∵α∈(0,π/2),∴sinα>0,cosα>0,sinα+cosα>
(2sina-3cosa)(sina+cosa)=0a是锐角,sina>0,cosa>0所以sina+cosa不等于0所以2sina=3cosasina=3/2*cosa代入sina+cosa=1co
sinα=4sin(α+β),sin(α+β-β)=4sin(α+β),sin(α+β)cosβ-cos(α+β)sinβ=4sin(α+β),sin(α+β)(cosβ-4)=cos(α+β)sin
运用公式sin(α+β)=sinα*cosβ+cosα*sinβsin(α-π/4)=sinα*cos(-π/4)+cosα*sin(-π/4)=(sinα-cosα)/√2=5/13(1)sinα^
∵sin(α-π4)=35,∴cos(α+π4)=cos[π2+(α-π4)]=-sin(α-π4)=-35.故答案为:-35
(1)由tan(α+π/4)=2得,tanα=tan[(α+π/4)-π/4]=(2-1)/(1+2×1)=1/3,所以,(2sinα+cosα)/(3cosα-2sinα)=(2tanα+1)/(3
∵sinα=23,且α是钝角,∴cosα=-1−sin2α=-1−(23)2=-53,tanα=sinα:cosα=23−53=-255,cotα=1tanα=-52.
有tan(π/4+α)=-3,可以得到tanα=2;从而右边分子分母同除以Cosα的平方得到tanα的关系式;故而求解得原试=28;
tan(α+π/4)=(tanα+tanπ/4)/(1-tanatanπ/4)=(tanα+1)/(1-tanα)=2所以tanα=1/32cosα-sinα/cosα+3sinα=(2-sina/c
sin(α-β)cosα-cos(β-α)sinα=sin(α-β-a)=-sin(β)=3/5=>sinβ=-3/5cosβ=-4/5sin(β+5π/4)=-sin(β+π/4)=-1/2*根号2