求函数y=(sinx+4)(cosx+4)的最小值

题目

求函数y=(sinx+4)(cosx+4)的最小值

答案

y=(sinx+4)(cosx+4)
=sinxcosx+4(cosx+sinx)+16
=(1/2)sin(2x) +4(cosx+sinx)+16
=(1/2)cos(π/2- 2x)+4(cosx+sinx)+16
=(1/2)cos[2(x-π/4)]+4√2cos(x-π/4)+16
=(1/2)[2cos²(x-π/4)-1]+4√2cos(x-π/4)+16
=cos²(x-π/4) +4√2cos(x-π/4)+31/2
=[cos(x-π/4) +2√2]²+15/2
当cos(x-π/4)=-1时,y有最小值
ymin=(2√2-1)²+15/2=23/2 -4√2