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Q-cos-1-1-2x-2-2sin-1-x-prove-please-

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Question Number 40466 by naziri2013@ last updated on 22/Jul/18
Q..cos^(−1) (1−2x^2 )=2sin^(−1) x,prove please
Q..cos−1(1−2x2)=2sin−1x,proveplease
Commented by maxmathsup by imad last updated on 22/Jul/18
let prove that arccos(1−2x^2 )=2arcsinx let f(x)=arccos(1−2x^2 )−2arcsinx D_f ? x∈ D_f ⇔ −1≤x≤1 and −1≤1−2x^2 ≤1 but −1≤1−2x^2 ≤1 ⇔−2≤−2x^2 ≤0 ⇔ 0≤2x^2 ≤2 ⇔0≤x^2 ≤1 ⇔ −1≤x≤1 ⇒ D_f =[−1,1] we have f^′ (x)= ((−(−4x))/( (√(1−(1−2x^2 )^2 )))) − (2/( (√(1−x^2 )))) =((4x(√(1−x^2 )) −2(√(1−(4x^4 −4x^2 +1))))/( (√(1−x^2 ))(√(1−(1−2x^2 )^2 )))) =((4x(√(1−x^2 )) −2(√(4x^2 −4x^4 )))/( (√(1−x^2 ))(√(4x^2 −4x^4 )))) = ((4x(√(1−x^2 )) −4∣x∣(√(1−x^2 )))/( (√(1−x^2 ))(√(4x^2 −4x^4 )))) if 0<x<1 we have f^′ (x)=0 ⇒f(x)=c ∀ x∈[0,1] f(x)=f(0)=0 if x∈]−1,0[ f^′ (x)= ((8x(√(1−x^2 )))/(−2x(1−x^2 ))) =((−4)/( (√(1−x^2 )))) ≠0 ⇒f is not constant finally we have arccos(1−2x^2 )=2 arcsinx if x∈[0,1] .
letprovethatarccos(1−2x2)=2arcsinxletf(x)=arccos(1−2x2)−2arcsinxDf?x∈Df⇔−1⩽x⩽1and−1⩽1−2x2⩽1but−1⩽1−2x2⩽1⇔−2⩽−2x2⩽0⇔0⩽2x2⩽2⇔0⩽x2⩽1⇔−1⩽x⩽1⇒Df=[−1,1]wehavef′(x)=−(−4x)1−(1−2x2)2−21−x2=4x1−x2−21−(4x4−4x2+1)1−x21−(1−2x2)2=4x1−x2−24x2−4x41−x24x2−4x4=4x1−x2−4∣x∣1−x21−x24x2−4x4if0<x<1wehavef′(x)=0⇒f(x)=c∀x∈[0,1]f(x)=f(0)=0ifx∈]−1,0[f′(x)=8x1−x2−2x(1−x2)=−41−x2≠0⇒fisnotconstantfinallywehavearccos(1−2x2)=2arcsinxifx∈[0,1].
Commented by prakash jain last updated on 22/Jul/18
This is not true in general range of cos^(−1) is [0,π] range of sin^(−1) is [−(π/2),(π/2)] put x=−(1/2) 2sin^(−1) x=−(π/3) cos^(−1) (1−2x^2 )=cos^(−1) (1/2)=(π/3)
Thisisnottrueingeneralrangeofcos−1is[0,π]rangeofsin−1is[−π2,π2]putx=−122sin−1x=−π3cos−1(1−2x2)=cos−112=π3
Commented by prakash jain last updated on 22/Jul/18
also note that sin^(−1) (sin x) does not always equal x due to restriction on range of sin^(−1) x. sin (sin^(−1) x) however always equals x.
alsonotethatsin−1(sinx)doesnotalwaysequalxduetorestrictiononrangeofsin−1x.sin(sin−1x)howeveralwaysequalsx.
Commented by prof Abdo imad last updated on 23/Jul/18
i have proved that equality is true in[0,1] and x=−(1/2) ∉[0,1] !
ihaveprovedthatequalityistruein[0,1]andx=−12∉[0,1]!
Commented by prakash jain last updated on 23/Jul/18
My comment was for the question. The questioner seems to be a student so wanted to highlight common mistakes made while using inverse trignimetric functions.
Mycommentwasforthequestion.Thequestionerseemstobeastudentsowantedtohighlightcommonmistakesmadewhileusinginversetrignimetricfunctions.
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jul/18
x=sinθ cos^(−1) (1−2sin^2 θ) cos^(−1) (cos2θ)=2θ 2sin^(−1) (sinθ) 2θ
x=sinθcos−1(1−2sin2θ)cos−1(cos2θ)=2θ2sin−1(sinθ)2θ

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