已知 sinα =1 5,且9π 2
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sin+cosα/sinα-cosα=21+tanα/tanα-1=2tanα=3α是第三象限角,因此sinα=-3根号10/10;cosα=-根号10/10p点的坐标为(cos(a-π/2).sin
∵a⊥b∴(sinα+cosα,√2sinα)*(cosα-sinα,√2cosα)=0(sinα+cosα)(cosα-sinα)+√2sinα√2cosα=0cos2α+sin2α=0tan2α=
证明:左减右得:1−tan 2α1+tan 2α−1−tan 2β2(1+tan 2β)=1−sin 2αcos 2α1+sin
解出来sin(α)根号二分之一,cos(β)=二分之根号三阿拉法是45°,β是30°
(1)∵向量a⊥向量b,∴向量a*向量b=0,即6(sinα)^2+5sinαcosα-4(cosα)^2=0,因式分解得(2sinα-cosα)(3sinα+4cosα)=0,∴tanα=1/2或-
1.f(α)=[sin(π-α)cos(2π-α)]/sin(3π-α)=sinacosa/sina=cosa2.α是第三象限角,sinα=-3/5,cosa=-4/5,f(α)=-4/53.f(α)
f(α)=sin(π-α)cos(2π-α)tan(-α+π)/sin(π+α)tan(2π-α)=sinacosa(-tana)/(-sina)(-tana)=-(sinacosatana/sina
(1)f(α)=[sin(π+α)cos(2π-α)tan(-α+3π/2)tan(-α-π)]/sin(π-α)=[(-sinα)*cosα*cotα*(-tanα)]/sinα=cosα;(2)∵
sinα-cosα=1/5,则有(sinα-cosα)^2=1/25即(sinα)^2+(cosα)^2-2sinαcosα=1/25sin2α=1-1/25=24/25cos2α=√(1-(sin2
sin(π-α)-cos(π-α)=sina+cosa=1/2-√3/2sina=1/2cosa=-√3/2tanα=-√3/3
cosα=-4/5原式=(1-cosα)/2+(2sin2αcos^22α)/2cos^22α=(1-cosα)/2+2sinαcosα=-3/50
∵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
1/sinβ=(cosαcosβ-sinαsinβ)sinα整理得:(1+cosα*cosα)sinβ=2sinαcosαcosβ所以,tanβ=sinαcosα/(1+1-2(sinα)^2)/2=
题目有问题...改:α、β为锐角,且3sin²α+2sin²β=1,3sin2α-2sin2β=0求证:α+2β=π/2.方法多,其一证明:由3sin²α+2sin&su
{2tanα+3sinβ=7}*24tanα+6sinβ=14tanα-6sinβ=1相加:tanα=3(tanα)平方=9=(sinα)平方/(cosα)平方=(sinα)平方/{1-(sinα)平
1.∵sinβ=sinαcos(α+β)∴sinβ=sinα(cosαcosβ-sinαsinβ)∴sinβ=sinαcosαcosβ-sin^2αsinβ∴sinβ(1+sin^2α)=sinαco