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已知函数F(x)=2Cosxsin(x+π3)?3sin2x+sinxCosx+2...

f(x)=?3sin2x+sinxcosx=?3×1?cos2x2+12sin2x=12sin2x+32cos2x?32=sin(2x+π3)?32,(I)T=2π2=π(II)∴0≤x≤π2,∴π3≤2x+π3≤4π3,∴?32≤sin(2x+

(1)∵sin25π6=12,cos25π6=32,∴f(25π6)=-3sin225π6+sin25π6cos25π6=0.(2)f(x)=32cos2x-32+12sin2x.∴f(α2)=32cosα+12sinα-32=14-3216sin2α-4sinα-11=0,解得sinα=1±358∵α∈(0,π),∴sinα>0故sinα=1+358.

2x+π/3∈(π/3,4π/3) 2x+π/3看成一个整体a.取值a∈(π/3,4π/3) sin a (-√3/2,1]就是个sin的函数. 所以sin(2x+π/3)∈(-√3/2,1]

f(x)=2cosxsin(x+π/3)-√3sin^2x+sinxcos =sin(x+π/3+x)+sin(x+π/3-x)-√3(1-cos^2x)+sinxcos =sin(x+π/3+x)+sin(π/3)-√3+√3cos^2x+0.5sin2x =sin(2x+π/3)+√3/2-√3+√3cos^2x+0.5sin2x =sin(2x+π/3)-√3/2+√3*(1+cos2x)/2+0.5sin2x =sin2x*co...

f(x)=2cosxcos(x一兀/6)一√3sin2x+sinxcosx =2cosx(√3/2cosx+1/2sinx)-√3sin2x+1/2sin2x =√3cos^2(x)+1/2sin2x-√3sin2x+1/2sin2x =√3cos^2(x) -√3sin2x+sin2x =√3/2(1+cos2x) -(√3-1)sin2x =√3/2+√3/2cos2x -(√3-1)sin2x 检查题目是否抄错?

稍候

f(x)=sin^2x+√3sinxcosx+2cos^2x =cos^2x+√3/2*2sinxcosx+1 =1/2cos2x+√3/2sin2x+3/2 =sinx(2x+π/6)+3/2 T=2π/2=π 2x+π/6在[2kπ-π/2,2kπ+π/2]上单调递增 x在[kπ-5π/12,kπ+π/6]上单调递增 2)y=sin2x的图像经X轴向左平移π/12个单位得到y=sinx(2x+...

(1)y=2-sin2x+cos2x=?2sin(2x?π4)+2,∵2kπ?π2≤2x?π4≤2kπ+π2,∴kπ?π8≤x≤kπ+38π,∴该函数在[0,π]上的单调递减区间为[0,38π],[7π8,π].(4分)(2)T=π,由(1)问知:当x=78π+kπ,(k∈Z),倍f(x)最大值为2+2,当x=38π+kπ,(k∈Z),f(x)...

y=2cosxsinx(x+π/3)-根号3sin^2x+sinxcosx =2cosx[1/2*sinx+√3/2*cosx]-√3sin^2x+sinxcosx =sinxcosx+√3cos^2x-√3sin^2x+sinxcosx =2sinxcosx+√3[cos^2x-sin^2x] =sin2x+√3cos2x =2(1/2sin2x+√3/2*cos2x) =2sin(2x+π/3) 周期:π 最值:±2 单调区...

f(x)=cos(2x+π/3)-cos2x+(√3)sin2x =cos(2x+π/3)-2[(1/2)cos2x-(√3/2)sin2x] =cos(2x+π/3)-2[cos2xcos(π/3)-sin2xsin(π/3)] =cos(2x+π/3)-2cos(2x+π/3)=-cos(2x+π/3) ①.单调区间: 由 2kπ≤2x+π/3≤2kπ+π,得单增区间为:[kπ-π/6,kπ+π/3]; 由 2kπ...

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