1、
f(x)=(√3/2)sin2x+(1/2)cos2x
=sin(2x+π/6)
周期T=2π/2=π
當(dāng)x∈【0,π/2】時(shí),
2x+π/6∈【π/6,7π/6】
則sin(2x+π/6)∈【-1/2,1】
所以,f(x)在區(qū)間【0,π/2】上的最大值為1,最小值為-1/2；
2、
即：sin(2x0+π/6)=5/13
x0∈【π/4,π/2】,則2x0+π/6∈【2π/3,7π/6】
所以,cos(2x0+π/6)=-12/13
cos2x0=cos[(2x0+π/6)-π/6]
=cos(2x0+π/6)cos(π/6)+sin(2x0+π/6)sin(π/6)
=(-12/13)(√3/2)+(5/13)(1/2)
=(5-12√3)/26