Question

已知函数fx=cos（x＋π/12）,gx=1＋1/2sin2x （Ⅰ）设x=x0是函数y=fx

Answer 1

(1)∵cos2x=2cos^2x-1

∴f(x)=1/2+cos(2x+π/6)/2

对称轴2x0+π/6=π+2kπ

x0=5π/12+kπ

g(x0)=1+1/2sin(5π/6+2kπ)=5/4

（2）h(x)=1/2+cos(2x+π/6)/2+1+1/2sin2x

=3/2+0.5*(√3/2*cos2x-1/2sin2x+sin2x)

=3/2+0.5*sin(2x+π/3)

递减区间为（π/12+kπ,7π/12+kπ）