Question

数学15题 求解过程

Answer 1

15.解：f(x)=cos[(6k+1)π/3+2x]+cos[(6k-1)π/3-2x]+2(√3)sin(π/3+2x)
=cos[2kπ+(π/3+2x)]+cos[(2kπ-(π/3+2x)]+2(√3)sin(π/3+2x)
=cos(π/3+2x)+cos(π/3+2x)+2(√3)sin(π/3+2x)
=2cos(π/3+2x)+2(√3)sin(π/3+2x)
=4[(1/2)cos(π/3+2x)+(√3/2)sin(π/3+2x)
=4[cos(π/3+2x)cos(π/3)+sin(π/3+2x)sin(π/3)]
=4cos[(π/3+2x)-(π/3)]=4cos2x.
故f(x)的最小正周期T=2π/2=π
f(x)的值域为[-4，4].

Answer 2

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