数学15题 求解过程
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发布时间:2024-01-18 01:24
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热心网友
时间:2024-08-10 17:28
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].
热心网友
时间:2024-08-10 17:35
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