【参考答案】
f(x)=cos(π/2 -x)+cosx
=sinx+cosx
=√2[(√2/2)sinx+(√2/2)cosx]
=√2sin(x+ π/4)
（1）最小正周期T=2π/1=2π,最大值是√2
（2）f(a)=3/4即
√2sin(a+ π/4)=3/4
sin(a+ π/4)=3√2/8
∴sin2a=sin[2(a+ π/4)-(π/2)]
=-cos[2(a+ π/4)]
=-[1-2sin²(a+ π/4)]
=2×(3√2/8)²-1
=-7/16
【参考答案】
f(x)=cos(π/2 -x)+cosx
=sinx+cosx
=√2[(√2/2)sinx+(√2/2)cosx]
=√2sin(x+ π/4)
（1）最小正周期T=2π/1=2π,最大值是√2
（2）f(a)=3/4即
√2sin(a+ π/4)=3/4
sin(a+ π/4)=3√2/8
∴sin2a=sin[2(a+ π/4)-(π/2)]
=-cos[2(a+ π/4)]
=-[1-2sin²(a+ π/4)]
=2×(3√2/8)²-1
=-7/16