Inverse Sine

Author

Eric W. Weisstein
August 24, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/SpecialFunctions/InverseSine.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/InverseSine.html.

©2008 Wolfram Research, Inc. except for portions noted otherwise

Definition

TrigToExp[ArcSin[x]]

Plots

<<Utilities`Plot`
<<Utilities`Typesetting`

Branch Cuts

BranchCuts[ArcSin]//TraditionalForm

With[{f=ArcSin},
BranchCutPlot[f,Epilog->Text[TraditionalForm[f[z]],{-2,2.3}]]
]

Plot[ArcSin[x],{x,-1,1},AxesLabel->TraditionalForm/@{x,ArcSin[x]},
PlotRange->{Automatic,{-1.8,1.8}},
PlotStyle->Red,
Epilog->{Red,PointSize[.02],Point/@{{-1,-Pi/2},{1,Pi/2}}},
Ticks->{Automatic,{#,TraditionalForm[#]}&/@Range[-Pi/2,Pi/2,Pi/4]}]

Complex Plane

ComplexPlot3D[ArcSin[z],{z,-5,5},PlotPoints->100,Mesh->False]//Timing

ComplexContourPlot[ArcSin[z],{z,-5,5}]//Timing

Special Values

ArcSin/@{-1,0,1}

FunctionExpand[ArcSin[#]]&/@{-I,I}

Derivative and Integral

Derivative

D[ArcSin[z],z]

Integral

Integrate[ArcSin[z],z]

Series

series=Series[ArcSin[x],{x,0,20}]

coefs=Last/@Partition[CoefficientList[series,x],2]

Through[{Numerator,Denominator}[coefs]]

SeriesCoefficient[ArcSin[x],{x,0,n}]

FullSimplify[%]

Sums

Sum[(((Subscript[((1/2)), ((n-1)/2)]) (x^n))/(n ((n-1)/2)!)),{n,1,∞,2}]==(sin^(-1))(x)

Sum[((Pochhammer[(1/2),(1/2) (-1+n)])/(n ((1/2) (-1+n))!))x^n,{n,1,∞,2}]

(∑_(n=0)^∞)(((Subscript[((1/2)), n]) (x^(2 n+1)))/((2 n+1) n!))==(sin^(-1))(x)

Sum[((Pochhammer[(1/2),n])/((2n+1)  n!))(x^(2n+1)),{n,0,∞}]

(∑_(n=0)^∞)(x^n) Piecewise[(({{((Subscript[((1/2)), ((n-1)/2)])/(n ((n-1)/2)!)), n mod 2==1∧n≥0}, {0, "PiecewiseDefault"[True]}}))]==(sin^(-1))(x)

Sum[x^n(Piecewise[{{((Pochhammer[(1/2),(1/2) (-1+n)])/(n ((1/2) (-1+n))!)), Mod[n,2]==1&&n≥0}, {0, True}}]),{n,0,∞}]

(1/2) (∑_(n=1)^∞)((((2 x))^(2 n))/((n^2) (({{2 n}, {n}}))))==(((sin^(-1))(x))^2)

V6

(1/2)Sum[((((2x))^(2n))/((n^2)Binomial[2n,n])),{n,∞}]

V7

(1/2)Sum[((((2x))^(2n))/((n^2)Binomial[2n,n])),{n,∞}]

(Underoverscript[(3/2)∑, (k=1), ∞])(((Subscript[H, k-1]^((2))) (x^(2 k)))/((({{2 k}, {k}})) (k^2)))==(((sin^(-1))((x/2)))^4)

V6

Sum[(1/(m^2)),{m,k-1}]//FullSimplify

Sum[HarmonicNumber[-1+k,2]((x^(2k))/(Binomial[2k,k](k^2))),{k,∞}]

V7

Sum[(1/(m^2)),{m,k-1}]//FullSimplify

(3/2)Sum[HarmonicNumber[-1+k,2]((x^(2k))/(Binomial[2k,k](k^2))),{k,∞}]

(45/4) (∑_k^∞)(((((((Subscript[H, k-1]^((2)))))^2)-(Subscript[H, k-1]^((4)))) (x^(2 k)))/(2 ((k^2) (Binomial[2 k, k]))))==(((sin^(-1))((x/2)))^6)

Sum[(1/(m^2))Sum[(1/(n^2)),{n,m-1}],{m,k-1}]

FullSimplify[%]

TimeConstrained[(45/4)Sum[(1/2) (((HarmonicNumber[-1+k,2])^2)-HarmonicNumber[-1+k,4])((x^(2k))/((k^2)Binomial[2k,k])),{k,∞}],60]//Timing

(45/4)Sum[(1/2) (((HarmonicNumber[-1+k,2])^2)-HarmonicNumber[-1+k,4])((x^(2k))/((k^2)Binomial[2k,k])),{k,100}]/.x->1.1

ArcSin[x/2]^6/.x->1.1

(315/2) (∑_k^∞)(((x^(2 k)) ((∑_m^(k-1))(((∑_n^(m-1))(((∑_p^(n-1))(1/(p^2)))/(n^2)))/(m^2))))/((k^2) (Binomial[2 k, k])))==(sin^8)((x/2))

Sum[(1/(n^2))Sum[(1/(p^2)),{p,n-1}],{n,m-1}]

TimeConstrained[Sum[(1/(m^2))(1/2) (((HarmonicNumber[-1+m,2])^2)-HarmonicNumber[-1+m,4]),{m,k-1}],120]//Timing

(315/2)Sum[Sum[(1/(m^2))Sum[(1/(n^2))Sum[(1/(p^2)),{p,n-1}],{n,m-1}],{m,k-1}]((x^(2k))/((k^2)Binomial[2k,k])),{k,∞}]

Identities

In Terms of Itself

For all z

FullSimplify[-ArcSin[-z]]

FullSimplify[ArcCos[-z]-Pi/2]

FullSimplify[Pi/2-ArcCos[z]]

FullSimplify[ArcCsc[1/x]]

In Terms of Other Trigs

FullSimplify[Pi/2-ArcCot[x/Sqrt[1-x^2]],x>0]

FullSimplify[-Pi/2-ArcCot[x/Sqrt[1-x^2]],x<0]

FullSimplify[Pi/2-ArcSin[Sqrt[1-x^2]],x>0]

FullSimplify[-Pi/2+ArcSin[Sqrt[1-x^2]],x<0]

For-1<x<1

FullSimplify[ArcTan[x/Sqrt[1-x^2]],-1<x<1]

FullSimplify[ArcCot[Sqrt[1-x^2]/x],-1<x<1]

FullSimplify[ArcCos[Sqrt[1-x^2]],-1<x<1]

FullSimplify[ArcCos[Sqrt[1-x^2]],-1<x<0]

FullSimplify[ArcCos[Sqrt[1-x^2]],0<x<1]

FullSimplify[ArcSec[1/Sqrt[1-x^2]],-1<x<1]

FullSimplify[ArcSec[1/Sqrt[1-x^2]],-1<x<0]

FullSimplify[ArcSec[1/Sqrt[1-x^2]],0<x<1]

Continued Fraction

With[{z=.2`10},{((z (Sqrt[1-(z^2)]))/(1+ContinuedFractionK[-Ceiling[n/2](Ceiling[n/2]+1)(z^2),2n+1,{n,30}])),ArcSin[z]}]//N

((z (Sqrt[1-(z^2)]))/(1+ContinuedFractionK[-Ceiling[n/2](Ceiling[n/2]+1)(z^2),2n+1,{n,∞}]))