Question

已知tana=-1/3,cosβ=(√5)/5,a,β∈(0,π)

Answer 1

(1)

sinβ=(1-(1/5))^(1/2)=2(√5)/5

tanβ=sinβ/cosβ=2

tan(a+β)=(tana+tanβ)/(1-tana*tanβ)

=1

(2)

tana=-1/3 a∈(π/2,π)

sina/cosa=-1/3

(sina)^2/(1-(sina)^2)=1/9

sina=(√10)/10

cosa=-3(√10)/10

f(x)=√2sin(x-a)+cos(x+β)

=√2(sinx*cosa-cosx*sina)+(cosx*cosβ-sinx*sinβ)

=-(√5)sinx