Question

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

需要过程 谢谢...

Answer 1

解：cos(β-π/4)=1/3
cos2(β-π/4)=2[cos(β-π/4)]^2-1=-7/9
cos(2β-π/2)=-7/9
cos(π/2-2β)=-7/9
sin2β=-7/9
cos(β-π/4)=1/3, sin(α+β)=4/5, 0＜α＜π/2＜β＜π
sin(β-π/4)=2√2/3
cos(α+β)=-3/5
cos(α+π/4)=cos[(α+β)+(π/4-β)]
=cos(α+β)cos(π/4-β)-sin(α+β)sin(π/4-β)
=cos(α+β)cos(β-π/4)+sin(α+β)sin(β-π/4)
=(8√2-3)/15