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F Cosx Cos2x

f(cosx) = cos2x = 2(cosx)^2-1 => f(x) = 2x^2 -1 f(sin15°)= 2(sin15°)^2 -1 = -(1-2(sin15°)^2) = - cos30° =-√3/2

f(sinx)=3-cos2x =3-(1-2sin²x) =2+2sin²x 则: f(x)=2+2x² 所以, f(cosx)=2+2cos²x =2+1+cos2x =3+cos2x

令t=sinx则 f(sinx)=f(t)=3-1+2sin²x=2+2t²,(-1≤t≤1) ∴f(t)=2+2t²,(-1≤t≤1) ∴f(cosx)=2+2cos²x=cos2x+3

解:设t=2-cosx∈[1,3]cosx=2-tcos2x=2(cosx)^2-1=2(t-2)^2-1=2t^2-8t+7故有:f(t)=2t^2-8t+7+2-t=2t^2-9t+9,t∈[1,3]f(x-1)=2(x-1)^2-9(x-1)+9=2x^2-4x+4-9x+9+9=2x^2-13x+22(x∈[2,4])

∵f(sinx)=3-cos2x=3-(1-2sinx*sinx) ∴f(x)=3-(1-2x*x) =2+2x*x 或者:设sinx=t cos2x=1-2(sinx)^2=1-2t^2 f(sinx)=3-cos2x f(t)=3-(1-2t^2)=2t^2+2 把t改成x f(x)=2x^2+2 f(cosx)=2cos^2 x+2=1+cos2x+2=3+cos2x

f(x)=(sinx+cosx)²+cos2x =sin²x+2sinxcosx+cos²x+cos2x =sin²x+cos²x+2sinxcosx+cos2x =1+2sinxcosx+cos2x =1+sin2x+cos2x =1+(sin2x+cos2x) =1+√2(√2/2sin2x+√2/2cos2x) =1+√2[cos(π/4)sin2x+sin(π/4)...

f(sinx)=cos2x=1-2sin²x 所以:f(x)=1-2x² 所以:f(cosx)=1-2cos²x=sin²x-cos²x=-cos2x

配凑法: f(2-cosx)=cos(2x)-cosx =2cos²x-1-cosx =2cos²x-8cosx+8+7cosx-14+5 =2(cos²x-4cosx+4)-7(2-cosx)+5 =2(2-cosx)²-7(2-cosx)+5 将2-cosx换成x f(x)=2x²-7x+5 f(x-1)=2(x-1)²-7(x-1)+5=2x²-4x+2-7x+...

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