高一三角函数f(x)=2cosxsin(x+π/3)-√3* (sin²x)+sinxcosx1.求函数f(x)的单调区间2.求函数f(x)图像的对称轴和对称中心3.求函数f(x)取得最大值时x的值
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高一三角函数f(x)=2cosxsin(x+π/3)-√3* (sin²x)+sinxcosx1.求函数f(x)的单调区间2.求函数f(x)图像的对称轴和对称中心3.求函数f(x)取得最大值时x的值
高一三角函数
f(x)=2cosxsin(x+π/3)-√3* (sin²x)+sinxcosx
1.求函数f(x)的单调区间
2.求函数f(x)图像的对称轴和对称中心
3.求函数f(x)取得最大值时x的值
高一三角函数f(x)=2cosxsin(x+π/3)-√3* (sin²x)+sinxcosx1.求函数f(x)的单调区间2.求函数f(x)图像的对称轴和对称中心3.求函数f(x)取得最大值时x的值
f(x)=2cosxsin(x+π/3)-√3* (sin²x)+sinxcosx
=2cosx(sinxcosπ/3+cosxsinπ/3)-√3* (sin²x)+sinxcosx
=2cosx(1/2sinx+√3/2cosx)-√3* (sin²x)+sinxcosx
=sinxcosx+√3(1-sin²x)-√3sin²x+sinxcosx
=2sinxcosx+√3(1-2sin²x)
=sin2x+√3cos2x
=2(1/2sin2x+√3/2cos2x)
=2(sin2xcosπ/3+cos2xsinπ/3)
=2sin(2x+π/3)
当 2x+π/3 ∈[2kπ-π/2,2kπ+π/2]时,f(x)单调增,此时 x∈[kπ-5π/12,kπ+π/12];
2x+π/3 ∈[2kπ+π/2,2kπ+3π/2]时,f(x)单调减,此时 x∈[kπ+π/12,kπ+7π/12];
(2) 由于 f(x)=sinx关于直线 kπ+π/2 对称,关于点(kπ,0)对称;
令
2x+π/3=kπ+π/2
得 x=kπ/2 +π/12 k∈Z;
所以原函数的对称轴为 x=kπ/2 +π/12 k∈Z;
令 2x+π/3=kπ,得 x=kπ/2 -π/6,k∈Z;
所以原函数的对称中心为 (kπ/2 -π/6,0)其中 k∈Z;
(3) f(x)=2sin(2x+π/3)
当 2x+π/3 =2kπ+π/2时,f(x)取得最大值2,此时
x=kπ+π/12.
教你七个字对待三角恒等变换:变角,变名,变次数(即统一名角和次数)
f(x)=2cosxsin(x+π/3)-√3* (sin2x)+sinxcosx=2sinxcosx +√3*cos2x=2sin(2x+π/3)
1)单调增区间为[kπ-5π/12, kπ+π/12] (k∈Z)
单调减区间为[kπ+π/12, kπ+7π/12] (k∈Z)
2)对称轴为 x= kπ/2+π/12, (k∈Z), 对称中心为(kπ/2 -π/6, 0),(k∈Z);
3)x=kπ+π/12, (k∈Z).
f(x)=2cosxsin(x+π/3)-√3* (sin²x)+sinxcosx=2sin(2x+π/3)
(1)当-π/2+2Kπ ≤2x+π/3≤π/2+2kπ,k∈Z,即kπ-5π/12≤x≤kπ+π/12,kπ+π/12,k∈Z时,f(x)单调递增
∴f(x)在[kπ-5π/12,kπ+π/12],k∈Z上单调递增;
当π/2+2kπ≤2x+π/3≤3π...
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f(x)=2cosxsin(x+π/3)-√3* (sin²x)+sinxcosx=2sin(2x+π/3)
(1)当-π/2+2Kπ ≤2x+π/3≤π/2+2kπ,k∈Z,即kπ-5π/12≤x≤kπ+π/12,kπ+π/12,k∈Z时,f(x)单调递增
∴f(x)在[kπ-5π/12,kπ+π/12],k∈Z上单调递增;
当π/2+2kπ≤2x+π/3≤3π/2+2kπ,k∈Z,即kπ+π/12≤x≤kπ+7π/12,k∈Z时,f(x)单调递减
∴f(x)在[kπ+π/12,kπ+7π/12]k∈Z上单调递减;
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