已知0<α<π⼀2<β<π,cos(β-π⼀4)=1⼀3,sin(α+β)=4⼀5 (1)求sin2β (2)求cos(α+π⼀4)的值

2024-07-11 23:24:39
推荐回答(1个)
回答1:

[cos(β-π/4)]^2=[cosβcos(π/4)+sinβsin(π/4)]^2
=[(cosβ)^2+(sinβ)^2]*[cos(π/4)]^2+2*cosβ*sinβ*cos(π/4)*sin(π/4)
=1/2+1/2*sin2β
=1/9
故 sin2β=-7/9

0<α<π/2<β<π 得 π/2〈α+β<3π/2 , π/4〈β-π/4〈3π/4;

cos(α+β)=-3/5 , sin(β-π/4)=2*1.414/3;

cos(α+π/4)=cos[(α+β)-(β-π/4)]
=cos(α+β)cos(β-π/4)+sin(α+β)sin(β-π/4)
带入即可,1.414为根号2。