已知函数f(x)根号3sinx/2cosx/2+cos^2x/2,求函数f(x)的最小正周期及单调递增区间
2个回答
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f(x)=(√3)[sin(x/2)][cos(x/2)]+cos²(x/2)?
是吗?
如果是的话:
最小正周期:2π/(1/2)=4π
单调递增区间是:x∈(4kπ-7π/6,4kπ+5π/6),其中:k∈N
解:
f(x)=(√3)[sin(x/2)][cos(x/2)]+cos²(x/2)
f(x)=cos(x/2)[(√3)sin(x/2)+cos(x/2)]
f(x)=2cos(x/2){[(√3)/2]sin(x/2)+(1/2)cos(x/2)}
f(x)=2cos(x/2)[cos(π/6)sin(x/2)+sin(π/6)cos(x/2)]
f(x)=2cos(x/2)sin(x/2+π/6)
f(x)=sin[(x/2+x/2+π/6)/2]cos[(x/2+π/6-x/2)/2]
f(x)=sin(x/2+π/12)cos(π/12)
最小正周期:2π/(1/2)=4π
f‘(x)=(1/2)cos(x/2+π/12)cos(π/12)
令:f'(x)>0,即:(1/2)cos(x/2+π/12)cos(π/12)>0
整理,有:cos(x/2+π/12)>0
得:2kπ-π/2<x/2+π/12<2kπ+π/2,其中:k∈N
即:4kπ-7π/6<x<4kπ+5π/6
f(x)的单调递增区间是:x∈(4kπ-7π/6,4kπ+5π/6),其中:k∈N
是吗?
如果是的话:
最小正周期:2π/(1/2)=4π
单调递增区间是:x∈(4kπ-7π/6,4kπ+5π/6),其中:k∈N
解:
f(x)=(√3)[sin(x/2)][cos(x/2)]+cos²(x/2)
f(x)=cos(x/2)[(√3)sin(x/2)+cos(x/2)]
f(x)=2cos(x/2){[(√3)/2]sin(x/2)+(1/2)cos(x/2)}
f(x)=2cos(x/2)[cos(π/6)sin(x/2)+sin(π/6)cos(x/2)]
f(x)=2cos(x/2)sin(x/2+π/6)
f(x)=sin[(x/2+x/2+π/6)/2]cos[(x/2+π/6-x/2)/2]
f(x)=sin(x/2+π/12)cos(π/12)
最小正周期:2π/(1/2)=4π
f‘(x)=(1/2)cos(x/2+π/12)cos(π/12)
令:f'(x)>0,即:(1/2)cos(x/2+π/12)cos(π/12)>0
整理,有:cos(x/2+π/12)>0
得:2kπ-π/2<x/2+π/12<2kπ+π/2,其中:k∈N
即:4kπ-7π/6<x<4kπ+5π/6
f(x)的单调递增区间是:x∈(4kπ-7π/6,4kπ+5π/6),其中:k∈N
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f(x)=√3sin(x/2)cos(x/2)+cos²(x/2)+1/2
=√3/2sinx+(cosx+1)/2+1/2
=√3/2sinx+1/2cosx+1
=sin(x+π/6)+1
令2kπ+π/2≤x+π/6≤2kπ+3π/2 (k∈N)
则2kπ+π/3≤x≤2kπ+4π/3 (k∈N)
∴f(x)的单调递减区间为[2kπ+π/3,2kπ+4π/3](k∈N)
=√3/2sinx+(cosx+1)/2+1/2
=√3/2sinx+1/2cosx+1
=sin(x+π/6)+1
令2kπ+π/2≤x+π/6≤2kπ+3π/2 (k∈N)
则2kπ+π/3≤x≤2kπ+4π/3 (k∈N)
∴f(x)的单调递减区间为[2kπ+π/3,2kπ+4π/3](k∈N)
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第一步的+1/2哪来的
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