已知f(x)=sinwx+sin(wx+π/2),w>0,且函数f(x)最小正周期为2π

sin(wx+π/2)=coswx
f(x)=sinwx+coswx
=√2(sinwx*√2/2+coswx*√2/2)
=√2sin(wx+π/4)
T=2π
∴w=1,f(x)最大值=√2
f(x)=√2sin(x+π/4)=sinx+cosx
f(a)=sina+cosa=3/4
sin²a+cos²a=1 (sina,cosa平方和=1)
(sina+cosa)²-2sinacos=1
2sinacosa=9/16-1=-7/16
sinacosa=-7/32
sina,cosa为二次方程x²-3x/4-7/32=0
(x-3/8)²=7/32+9/64=23/64
x=(3±√23)/8
a属于(0,π),∴cosa>0,舍负根,负根为sina的值
∴cosa=(3+√23)/8