Inverse Trigonometric Functions

Supporting materials

If you wish to get a different perspective on the notes below try either of the following textbook sections.

Stewart’s Calculus
	Section 6.6: Inverse trigonometric functions

Boelkins/Austin/Schlicker’s Active Calculus
	Section 2.6: Derivatives of inverse functions

Sine function

The graph of the sine function f(x)=sin(x) is

Plot[Sin[x],{x,-10,10}]

Questions

Solve sin(x)=0 for x.

Solve sin(x)=1 for x.

Solve[Sin[x]==1,x]

Solve sin(x)=2 for x. (Video)

What is the domain and range of sin(x)?

Is sin(x) a one-to-one function over its entire domain? (Video)

Inverse sine

The function f(x)=sin(x) for -π/2≤x≤π/2 is a one-to-one function and has an inverse called the inverse sine or arc sine function
	(f^(-1))(x)=(sin^(-1))(x)=arcsin(x)(Video)

Plot[{ArcSin[x],x,Sin[x]},{x,-π/2,π/2},PlotLabel->HoldForm[y=ArcSin[x]]]

Note that the inverse sine is odd, just as sine is odd.

Questions

What are the domain and range of (sin^(-1))(x)? (Video)

What is (sin^(-1))(1)?(Video)

What is sin((sin^(-1))(0.2))?

For what values of x does (sin^(-1))(sin(x))=x?

Plot[ArcSin[Sin[x]],{x,-2Pi,2Pi}]
(* Mathematica Occasionally Lies, sort of: *)
Plot[Sin[ArcSin[x]],{x,-2Pi,2Pi}]
Plot[ArcSin[x],{x,-2Pi,2Pi}]

What is (sin^(-1))(sin(2π))? (Video)

ArcSin[Sin[2 Pi]]

Sketch the graph y=(sin^(-1))(-x) onto the graph above.

Solve 4(sin^(-1))(3x)=π for x. (Video)

Solve[4 ArcSin[3 x]==Pi,x]

Derivative of inverse sine

To find the derivative of (sin^(-1))(x), set y=(sin^(-1))(x). This is equivalent to
	sin(y)=x
Find ((d y)/(d x)) using implicit differentiation.
	(d/(d x))sin(y)=(d/(d x))x
	⟶ cos(y)((d y)/(d x))=1
	⟶ ((d y)/(d x))=(1/(cos(y)))

We need the answer in terms of x, not y. Since y=(sin^(-1))(x), we could write the derivative as
	((d y)/(d x))=(1/(cos((sin^(-1))(x))))
but there is a preferred form. Since -π/2≤y≤π/2, angle y is in the first or fourth quadrant; cos(y)≥0. Using this along with the Pythagorean identity
	(cos^2)(y)+(sin^2)(y)=1
to write cos(y) in terms of sin(y).

D[ArcSin[x],x]

Questions

Write cos(y) in terms of sin(y) if -π/2≤y≤π/2. (Video)

Using x=sin(y), write ((d y)/(d x))=(1/(cos(y))) in terms of x. (Video)

Derivative formula

(d/(d x))((sin^(-1))(x))=(1/(Sqrt[1-(x^2)]))

Questions

Find an equation for the tangent line to y=x (sin^(-1))(x) at x=1/2.

f[x_]:=x ArcSin[x]
x0=1/2
f'[x0](x-x0)+f[x0]
Show[
Plot[{f[x],f'[x0](x-x0)+f[x0]},{x,0,1}],
ListPlot[{{x0,f[x0]}}]
]

Let g(x)=(sin^(-1))(x/2)-x.

What is the domain of g(x)?

Find the maximum and minimum values of g(x).

What is the range of g(x)? (Video)

g[x_]:=ArcSin[x/2]-x
g'[x]
Plot[g[x],{x,-2,2}]
Solve[g'[x]==0,x]

What is ∫(1/(Sqrt[1-(x^2)]))d x?

Integrate[1/Sqrt[1-x^2],x]

What is ∫(1/(Sqrt[1-(x^2)]))d x? (Video)

Find the area bounded by y=(1/(Sqrt[1-(x^2)])), the x-axis, and x=±1/2.

Integrate[1/Sqrt[1-x^2],{x,-1/2,1/2}]

Evaluate the definite integral (∫_(ln(1/2))^(ln(1/(Sqrt[2]))))((e^t)/(Sqrt[1-(e^(2t))]))d t

Integrate[1/Sqrt[1-x^2],{x,1/2,1/Sqrt[2]}]

Inverse tangent

The graph of the periodic function y=tan(x) is

Plot[Tan[x],{x,-6,6},Exclusions->{-3π/2,-π/2,π/2,3π/2},PlotLabel->HoldForm[y=Tan[x]]]
Plot[{Tan[x],ArcTan[x]},{x,-2Pi,2Pi},Exclusions->{-3π/2,-π/2,π/2,3π/2},PlotLabel->HoldForm[Tan[x] and ArcTan[x]]]

Inverse tangent

The function h(x)=tan(x), restricted to -π/2<x<π/2, is a one-to-one function and has an inverse called the inverse tangent or arc tangent function
	(h^(-1))(x)=(tan^(-1))(x)=arctan(x)

Show[
Plot[{ArcTan[x],x,-Pi/2,Pi/2},{x,-4,4},PlotRange->{-4,4},PlotLabel->HoldForm[y=ArcTan[x]],AspectRatio->Full,PlotStyle->{Black,Dashed,{Thin,Black,Dashed},{Thin,Black,Dashed}},PlotLabels->Placed[Automatic,Above]],
Plot[{Tan[x]},{x,-π/2,π/2},PlotStyle->Dashed,PlotLabels->Placed[Automatic,Above]]
]

The inverse tangent function is symmetric (odd), “sigmoidal” -- s-shaped -- and asymptotes from -π/2 to π/2.

Questions

Find the solutions closest to 0 to the equation 3(tan^3)(x)+1=0.(Video)

ArcTan[-(1/3)^(1/3)]
N[%]

Since tan and (tan^(-1)) are inverse functions, y=(tan^(-1))(x) ⟷ tan(y)=x. Use this along with implicit differentiation to determine the derivative of (tan^(-1))(x), 		((d y)/(d x))=?(Video)

D[ArcTan[x],x]

What is the new antidifferentiation formula we can get from this derivative? (Video)

Inverse cosine

The graph of y=cos(x) is

Show[
Plot[Cos[x],{x,-6,6},PlotLabel->HoldForm[y=Cos[x]]],
Plot[Cos[x],{x,0,Pi},PlotStyle->{Thick,Black}]
]

Question

What are the domain and range of cos(x) (Video)?

Inverse cosine

The function k(x)=cos(x), restricted to 0≤x≤π (the bold black section of the curve above), is a one-to-one function and has an inverse called the inverse cosine or arc cosine function, denoted
	(k^(-1))(x)=(cos^(-1))(x)=arccos(x)

Plot[{ArcCos[x],x,If[x>0,Cos[x]]},{x,-1,π},PlotLabel->HoldForm[y=ArcCos[x]],AspectRatio->Full,
PlotStyle->{Black,Dashed,{Blue,Dashed}},PlotLabels->Placed[Automatic,Above]]

The inverse cosine is quite elegant, I think! But it lacks the obvious symmetry of the arcsin or arctan.

Questions

Since cos and (cos^(-1)) are inverse functions, y=(cos^(-1))(x) ⟷ cos(y)=x. Use this along with implicit differentiation to determine the derivative of (cos^(-1))(x), ((d y)/(d x))=?(Video)

D[ArcCos[x],x]

How does this compare to the derivative of (sin^(-1))(x)? (Video)

Summary

Inverse sine

The domain of (sin^(-1))(x) is [-1,1].

The range of (sin^(-1))(x) is [-π/2,π/2].

(d/(d x))((sin^(-1))(x))=(1/(Sqrt[1-(x^2)]))

Inverse tangent

The domain of (tan^(-1))(x) is (-∞,∞).

The range of (tan^(-1))(x) is (-π/2,π/2).

(d/(d x))((tan^(-1))(x))=(1/(1+(x^2)))

Inverse cosine

The domain of (cos^(-1))(x) is [-1,1].

The range of (cos^(-1))(x) is [0,π].

(d/(d x))((cos^(-1))(x))=-(1/(Sqrt[1-(x^2)]))

Integrals

∫(1/(Sqrt[1-(x^2)]))d x={({{(sin^(-1))(x)+(Subscript[C, 1])}, {or}, {-(cos^(-1))(x)+(Subscript[C, 2])}})

∫(1/(1+(x^2)))d x=(tan^(-1))(x)+C

Homework

IMath problems on inverse trigonometric functions.

Weekly assignment 2 due Wednesday, January 27.