已知α属于 (π/2,π),sinα=根号5/5，(1)求sin(π/4+α)(2)求cos(5π

/6-2α)

解：∵α∈(π/2,π)
∴cosα<0
∵sinα=√5/5
∴cosα=-√[1-(sinα)^2]=-2√5/5
故（1）sin(α+π/4)=sinαcos(π/4)+cosαsin(π/4)
=(√5/5)(√2/2)+(-2√5/5)(√2/2)
=-√10/10
（2）cos(5π/6-2α)=cos[π-(π/6+2α)]
=-cos(π/6+2α)
=-cos(π/6)cos(2α)+sin(π/6)sin(2α)
=-(√3/2)[(cosα)^2-(sinα)^2]+(1/2)(2sinαcosα)
=-(√3/2)[(-2√5/5)^2-(√5/5)^2]+(1/2)[2(√5/5)(-2√5/5)]
=3√3/10-2/5。

①

sina+cosa=1/5

∴(sina+cosa)²=sin²a+cos²a+2sinacosa=1+2sinacosa=1/25

∴2sinacosa=-24/25

∴(cosa-sina)²=sin²a+cos²a-2sinacosa=1-2sinacosa=1-(-24/25)=49/25

∵π/2<a<π

∴cosa<0,sina>0, 即cosa-sina<0

∴cosa-sina=-7/5

②b=π/3,cosa=4/5

∴sina=3/5,sinb=√3/2,cosb=1/2

∴sinc=sin(π-b-a)=sin(a+b)=sinacosb+sinbcosa=(3/5)×(1/2)+(√3/2)×(4/5)=(3+4√3)/10