f(x)=sinx+cos(x-π/3)的值域是多少
f(x)=sinx+cos(x-π/3)
=sinx+cosxcos(π/3)+sinxsin(π/3)
=(1+√3/2)·sinx+1/2·cosx
=(√6+√2)/2·[(√6+√2)/4·sinx+(√6-√2)/4·cosx]
=(√6+√2)/2·[cos75°·sinx+cos75°·cosx]
=(√6+√2)/2·sin(x+75°)
故:-(√6+√2)/2≤f(x)=sinx+cos(x-π/3)≤(√6+√2)/2
即:值域为:[-(√6+√2)/2,(√6+√2)/2]
注:(1+√3/2)²+(1/2)²=(8+4√3)/4=[(√6+√2)/2]²
热门标签: