Section 6.6: Inverse Trigonometric 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)=2 for x.

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

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

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)

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

Questions

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

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

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))?

ArcSin[Sin[2 Pi]]

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

Solve 4(sin^(-1))(3x)=π for x.

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.

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

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

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

Questions

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)?

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]

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 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]]]

Questions

What are the domain and range of tan(x)?

What is (lim_(x->π/(SuperMinus[2])))tan(x)?

What do you think is the standard restricted domain used to define an inverse tangent function, (tan^(-1))(x) or arctan(x)?

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

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))=?

D[ArcTan[x],x]

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

Inverse cosine

The graph of y=cos(x) is

Plot[Cos[x],{x,-6,6},PlotLabel->HoldForm[y=Cos[x]]]

Questions

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

What do you think is the standard restricted domain used to define an inverse cosine function, (cos^(-1))(x) or arccos(x)?

Plot[{ArcCos[x],Cos[x]},{x,-1,Pi}]

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))=?

D[ArcCos[x],x]

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

Inverse secant

The graph of y=sec(x) is

Plot[Sec[x],{x,-6,6},PlotLabel->HoldForm[y=Sec[x]]]

Questions

What are the domain and range of sec(x)?

There are two “standard” restricted domain used to define an inverse secant function, (sec^(-1))(x) or arcsec(x). What do you think they are?

Plot[{ArcSec[x],Sec[x]},{x,-Pi,Pi}]

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

D[ArcSec[x],x]