...α属于 (π/2,π),sinα=根号5/5,(1)求sin(π/4+α)(2)求cos(5π
发布网友
发布时间：2024-10-02 13:56
共1个回答
热心网友
时间：2024-10-19 07:07
∴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。
