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y=sin²x+2sinxCosx+3Cos²x化简

把这个函数化简就OK f(x)=sin²x+2sinxcosx+3cos²x =1+sin2x+2cos²x =1+sin2x+1+cos2x =2+√2sin(2x+π/4) 函数最大值=2+√2,最小值=2—√2 最小正周期=π 化简当中,如果看不懂的,可以问我。

f(x)=2cosx·sin(x+π/3)-√3sin²x+sinx·cosx =2cosx·(sinx·cosπ/3+sinπ/3·cosx)-√3sin²x+sinx·cosx =cosx·sinx+√3cos²x-√3sin²+sinx·cosx =√3(cos²x-sin²x)+2·sinx·cosx =√3cos2x-sin2x =2sin(2x+π/3) ①最小...

y=(sin²x+cos²x)/sin²x+2((sin²x+cos²x)/cos²x =sin²x/sin²x+cos²x/sin²x+2sin²x/cos²x+2cos²x/cos²x =3+cos²x/sin²x+2sin²x/cos²x≥3+2√(cos²...

解:y=7-4sinxcosx+4cos²x-4cos⁴x =7-2sin2x+4cos²x(1-cos²x) 【公式:2sinxcosx=sin2x】 =7-2sin2x+4cos²xsin²x =7-2sin2x+(2sinxcosx)² =7-2sin2x+sin²2x =6-(1-sin2x)² ∵-1≤sin2x≤1 ∴y...

cosπ/4cosx-sinπ/4sinx=3/5 cosx-sinx=3√2/5 平方 cos²x+sin²x-2sinxcosx=18/25 1-2sinxcosx=18/25 sinxcosx=7/50 sinx+cosx=√2sin(x+π/4) 5π/4

f(x)=5cos²x+sin²x-4√3sinxcosx =1+4cos²x-2√3sin(2x) =1+2+2cos(2x)-2√3sin(2x) =3-4[√3sin(2x)/2-cos(2x)/2] =3-4[sin(2x)cos(π/6)-cos(2x)cos(π/6)] =3-4sin(2x-π/6) =3-4sin[2(x-π/12)] T=2π/2=π f(x)的最小正周期是π 2x-π/6...

解:令t=sinx+cosx=√2sin(x+π/4) 故:-√2≤t≤√2 故:t²=1+2sinxcosx 故:sinxcosx=(t²-1)/2 故:f(x)=sinx+cosx+sinxcosx =t+(t²-1)/2 =1/2•(t²+2t-1) =1/2•(t+1)²-1 当t=-1时,函数f(x)=sinx+cosx...

sinx=sin2x ,sinx=2sinxcosx,sinx(1-2cosx)=0,sinx=0或者cosx=1/2,当sinx=0,x=kπ,当cosx=1/2,x=2kπ+π/3或者x=2kπ-π/3 sinx=cos2x,sinx=1-2sin²x,2sin²x+sinx-1=0,(2sinx-1)(sinx+1)=0,sinx=-1或者sinx=1/2,当sinx=-1,x...

原式=d/dx∫(0→cosx)cos(πt²)dt-d/dx∫(0→sinx)cos(πt²)dt =d/dcosx∫(0→cosx)cos(πt²)dt·dcosx/dx-d/dsinx∫(0→sinx)cos(πt²)dt·dsinx/dx =cos(πcos²x)(-sinx)-cos(πsin²x)cosx =-sinx·cos(πcos²x)-cosx·cos(πs...

解: 设M=sinx+cosx 则M=√2[√2/2sinx+√2/2cosx] =√2[cos(π/4)sinx+sin(π/4)cosx] =√2sin(x+π/4) ∴M的值域为[-√2,√2] 又∵M²=(sinx+cosx)²=sin²x+cos²x+2sinxcosx=1+2sinxcosx ∴sinxcosx=(M²-1)/2 ∴y=(M²-1)/2+M=(...

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