Question

已知0<α<π/2<β<π，cos(β－π/4）=1/3,sin(α-β)=4/5

1,求sin2β的值 2，求cos(α+π/4)的值

Answer 1

0<α<π/2<β<π，
∴-π<α-β<0,π/4<β-π/4<3π/4,
cos(β－π/4）=1/3,sin(α-β)=4/5,
∴sin(β-π/4)=2√2/3，cos(α-β)=土3/5.
（1）cosβ=cos(β-π/4+π/4)=1/3*1/√2-2√2/3*1/√2=（√2-4）/6，
sinβ=√[36-(√2-4）^2]/6=(4+√2)/6,
∴sin2β=2sinβcosβ=-7/9.
（2）cos(α+π/4)=cos[(α-β)+(β-π/4)+π/2]
=-sin[(α-β)+(β-π/4)]
=-[4/5*1/3土3/5*2√2/3]
=-(4土6√2）/15.