cos(π/4+a)=cos[π/2-(π/4-a)]=sin(π/4-a)
∵cos(π/4+a)cos(π/4-a)=1/4
∴ sin(π/4-a)cos(π/4-a)=1/4
(1/2)sin(π/2-2a)=1/4
∴ (1/2)cos(2a)=1/4
cos(2a)=1/2
sin²(2a)=1-cos²(2a)=1-1/4=3/4
sina的四次方加cosa的四次方
=(sin²a+cos²a)-2sin²acos²a
=1-(1/2)(2sinacosa)²
=1-(1/2)(sin2a)²
=1-(1/2)*3/4
=1-3/8
=5/8