Question

这些题怎么做呀？

1.cosα=1∕3，α∈（3π∕2,2π），求sin( π/3+α).
2.若cosθ=(-5/13),且θ∈( π/2, π).求cos2θ与tan2θ.3.
3.已知0<α< π/2,- π /2<β<0,sinα=4√3/7,sinβ=-11/14,求α+β.
做好了就给20分

Answer 1

1.α∈（3π∕2,2π），cosa=1/3,则sina=-2倍根号2/3,
sin(π/3+a)=sin(π/3)cosa+cos(π/3)sina
=(根号3/2)*1/3+1/2*（-2倍根号2/3）=（根号3-2倍根号2）/6
2.θ∈( π/2, π).cosθ=(-5/13),则sinθ=12/13
cos2θ=cos²θ-sin²θ=2cos²θ-1=-119/169
tanθ=sinθ/cosθ=-12/5
tan2θ=2tanθ/(1-tan²θ)=-120/119
3..已知0<α< π/2,- π /2<β<0,sinα=4√3/7,sinβ=-11/14,
则有，cosα=1/7,cosβ=5倍根号3/14
cos(α+β)=coscosβ+sinαsinβ=根号3/2
∵0<α< π/2,- π /2<β<0,∴-π/2<α+β<π/2
∴α+β为π/6或-π/6