InverseTrig.html

Inverse Trigonometric Functions

> restart:with(plots):

Inverse Functions in General

Definition : f is a one-to-one (1-1) function if whenever it must be that .

e.g. is not 1-1 since f(-2) = f(2) but ! Notice: fails the horizontal line test : f is 1-1 if and only if every horizontal line that crosses the graph of y = f(x) crosses it at only one point.

e.g. Show that f(x) = is 1-1.

> f:=x->(2-x)/(x+4);

> f(a);f(b);

> simplify(f(a)=f(b));

> solve(f(a)=f(b),a);

=> f is 1-1. WHY??

Check out the plot. Does it pass the horizontal line test??

> plot(f(x),x=-15..15,y=-4..4,discont=true,colour=red,thickness=2);

Definition : If f is a 1-1 function with domain D(f) and range R(f) then there exists a function g with domain R(f) and range D(f) that satisfies g(f(x)) = x for all x D(f) and f(g(x)) = x for all x R(f). g is called f-inverse and denoted .

It follows from the above definition that the range of f is the domain of and that the range of is the domain of f .

Find if f(x) = .

Note: we have already shown that f is 1-1 in the previous example. Therefore we know that its inverse exists. Now, suppose g is f-inverse and let a = g(x). It follows, by definition , that f(a) = x. Solve this equation for a!!

> f(a);

> solve(f(a)=x,a);

> g:=x->(2-4*x)/(1+x);

> f(5);

> g(-1/3);

> g(-13);

> f(-9/2);

> f(g(x));

> simplify(%);

> g(f(x));

> simplify(%);

> plot({f(x),g(x),x},x,y=-8..8,discont=true,colour=[black,red,navy],thickness=2);

e.g. Determine if .

> f:=x->m*x+b;

> solve(f(a)=f(c),a);

The above line makes it look like f is 1-1 but Maple is assuming something that is not necessarily true . Find it!!

Now, let a = g(x) and solve f(a) = x for a:

> solve(f(a)=x,a);

> g:=x->(1/m)*x-(b/m);

Let's check:

> f(g(x));g(f(x));

Now, make a general statement (theorem).

Inverse Trigonometric Functions

The Inverse Sine function.

The sine function is so famous and so important that it demands an inverse. However, if you look at the graph of y = sin(x) you see right away that it is not 1-1 and so has no inverse!

> plot(sin(x),x=-20..20,colour=blue,thickness=2);

You can see right away that sin(a) = sin(b) does not imply that a = b. For example: sin( ) = sin( ) but .

To solve this problem of non 1-1'ness we simply erase almost all of the graph! That is, we restrict the domain of f(x) = sin(x) to the interval on which the function is 1-1 and so has an inverse. We simply make sure to define the inverse g in such a way that it satisfies f(g(x)) = g(f(x)) = x. We call g(x), or "sine inverse of x". See blackboard for notation.

Definition : <=> for y and x .

*** Check that the above definition satisfies the composition requirements for inverse functions.

> sin(arcsin(x));arcsin(sin(x));

> [arcsin(-1),arcsin(0),arcsin(1/2),arcsin(1/sqrt(2)),arcsin(1)];

Simplify i) cos(arcsin(2/7)) ii) cos(arcsin(x))

> cos(arcsin(2/7));cos(arcsin(x));

> plot(arcsin(x),x=-Pi/2..Pi/2,colour=red,thickness=2);

> plot({sin(x),arcsin(x),x},x=-Pi/2..Pi/2,colour=[blue,red,magenta],thickness=3);

e.g. Find the derivative of y = arcsin(x) and generalize.

> d:=diff(arcsin(x),x);

Theorem : . Proof: Done by implicit differentiation . See Blackboard.

Chain rule generalization: OBVIOUS!

> eg1:=arcsin(x^2)+(arcsin(x))^2;

> diff(eg1,x);

The Others.

Arccos(x)

> plot(cos(x),x=-20..20,colour=red,thickness=2);

To solve the non 1-1'ness of y = cos(x) we restrict the domain to and say:

definition : <=> for y and x [-1,1] .

You should check that the definition satisfies the composition equations.

> arccos(-1);arccos(0);arccos(1/2);arccos(1);

Simplify arccos(tan(3/a) if a > 0.

> assume(a>0);tan(arccos(3/a));

> simplify(%);

> plot({arccos(x),cos(x),x},x=-1..Pi,colour=[black,red,magenta],numpoints=1500,thickness=3);

Arctan(x)

> plot(tan(x),x=-10..10,y=-10..10,discont=true,colour=red,thickness=2);

To solve the non 1-1'ness of y = tan(x), restrict the domain to and say:

definition : artan(x) = y <=> x = tan(y) for y and x R .

Simplify i) sec(arctan(5/3)) ii) sec(arctan(x/3)).

> sec(arctan(5/3));sec(arctan(x/3));

> plot({arctan(x),Pi/2,-Pi/2},x=-10..10,y=-2..2,thickness=2);

> Limit(arctan(x),x=infinity)=limit(arctan(x),x=infinity);Limit(arctan(x),x=-infinity)=limit(arctan(x),x=-infinity);

Now, use implicit differentiation to find the derivatives of the arcosine and arctangent functions.

> Diff(arccos(x),x)=diff(arccos(x),x);Diff(arctan(x),x)=diff(arctan(x),x);

Differentiate the following expressions:

> e1:=arctan(x^3);e2:=arccos(sqrt(2*t-1));e3:=(arctan(x))^(-1);

>

> diff(e1,x);diff(e2,t);diff(e3,x);

#65, pg 234. A ladder 10 ft long leans against a vertical wall. If the bottom of the ladder slides away from the base of the wall at a speed of 2 ft/s, how fast is the angle between the ladder and the wall changing when the bottom of the ladder is 6 ft from the base of the wall? (To solve this problem, an example of a "related rates" problem, you have to realize that both variables are also functions of time and an equation relating both of them can be differentiated implicitly with repect to the "hidden" variable t .)

> equ:=cos(alpha)=x/10;