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(2Cos²x%1)/2tAn(π/4%x)sin²(π/4+x)

tanx = 1/2 2sinx= cosx 2sinxcosx = (cosx)^2 sin2x = 1/(secx)^2 = 1/[( tanx)^2 +1] =1/(1+ 1/4) = 4/5 [sin(π/4+x)]^2 =[(√2/2)(sinx+cosx) ]^2 =(1/2)(1+sin2x) =(1/2)(1+ 4/5) = 9/10

tan(π/4-x)= (tanπ/4-tanx)/(1+tanπ/4tanx) = (1-tanx)(1+tanx) = (cosx-sinx)/(cosx+sinx) sin^2(π/4+x) = (√2/2sinx+√2/2cosx)^2= 1/2(sinx+cosx)^2 f(x) = [2cos^4x-2cos^2x+1/2]/[tan(π/4-x)sin^2(π/4+x)] =[2cos^4x-2cos^2x+1/2]/[ (...

tanx = 1/2 2sinx= cosx 2sinxcosx = (cosx)^2 sin2x = 1/(secx)^2 = 1/[( tanx)^2 +1] =1/(1+ 1/4) = 4/5 [sin(π/4+x)]^2 =[(√2/2)(sinx+cosx) ]^2 =(1/2)(1+sin2x) =(1/2)(1+ 4/5) = 9/10

解:f(x)=x²sinα/3+x²cosα/2+tanα f'(x)=2xsinα/3+2xcosα/2=2x(sinα/3+cosα/2)=2x*√(1/9+1/4)*sin(α+arctan3/2)=√13/3*xsin(α+arctan3/2) f'(1)=√13/3*sin(α+arctan3/2) α范围是[0,5π/12],故α+arctan(3/2)范围是[arctan3/2,5π/...

令tan(x-π/4)=0 x-π/4=kπ x=π/4+kπ 在(0,2π) x可取,π/4,5π/4, 则,间断点,两个

-π/2

f(x)=tan(3x+π/4) 1。f(π/9)=tan(π/3+π/4) =(tanπ/3+tanπ/4)/(1-tanπ/3tanπ/4) =(√3+1)/(1-√3) =-√3-2 2. ∵f(a/3+π/4)=2 ∴tan(a+3π/4+π/4) = tan(π+a)=tana=2 sina/cosa=2 sina=2cosa代入sin²a+cos²a=1 cos²a=1/5 ∴cos2a=2cos...

tan(π/4+x)=3,tan²(π/4+x)=9,cos²(π/4+x)=1/10,sin²(π/4+x)=9/10,cos(π/2+2x)=2*1/10-1=-4/5,sin2x=4/5 ,sin²2x=16/25,4sin²x*cos²x=16/25,sin²x*cos²x=4/25,(1-cos²x)*cos²x=4/25, sin...

分母=2tan(π/4-x)*cos^2[π/2-(π/4+x)] =2sin(π/4-x)cos(π/4-x) =sin(π/2-2x) =cos2x 分子=1/2[4cos^4(x)-4cos^2(x)+1] =1/2(cos2x)^2 所以原式=1/2(cos2x)^2/cos2x=1/2*cos2x

-√6/3 根据公式:(cosx-sinx)²=1-2sinxcosx=1-1/3=2/3 所以 cosx-sinx=+-√6/3 因为 π/4

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