JT's AP Calculus (AB)

Activity #08:   Trig  Inverses

John Tinnin                                        Last Modified:  09/08/03

Purpose

What are we about to do?

Objective:  Students will
1)  Investigate the inverse trig functions and determine their characteristics.
2)  Use the inverse trig functions in solving equations used to solve application problems
3)  Investigate the inverse exponenential (Logarithmic) functions and determine their characteristics.

Why are we doing this?

How did this develop?

Explanation & Examples

The Inverse of Sin[x]

Introduction

Let's see what all this means with trig functions.  Suppose we have the equation y = Sin[x] and we are given that y = and we want to know x.  The equation then reads = Sin[x] and we want to solve it for x.  We know now, that to solve it you simply enter into your calculator and hit INV  - SIN.  Depending on what mode the calculator is in, the display will either say 30 (if in degrees) or .5235988 (if in radians).  Thus you solved the equation for x.  The key stroke INV-SIN invokes a call to the inverse function for sine.  This function is called the ArcSine and is either denoted as ArcSin[x] or [x].  This notation and the use of the word ArcSine isn't new, we have used it before, but now we will look at this in a much more closely.  

The graph of y = Sin[x] over [-2π,2π] is:

We know that this graph tells us that to find the sine of any angle in that interval all we need to do is find our angle on the horizontal axis and draw a perpendicular line to the axis at that point.  We then move up (or down) that perpendicular to the point of intersection with the graph and draw a perpendicular line to the y axis from that point.  The point of intersection of this line with the y axis gives us the sine of the angle with measure x.  This is demonstrated in the graph below.

We could use this same graph however, to go the opposite direction.  Suppose we are given a number between -1 and 1 inclusive, which corresponds to the sine of an angle.  We wish to know what angle will give that sine.  From that point on the y axis we draw a perpendicular to the y axis at that point.  Move along that line until you reach a point of intersection with the graph and drop a perpendicular to the x axis.  The intersection of that perpendicular with the x axis gives the angle which has the given sine.
From the picture one can see immediately that we have a real serious problem.  When we draw the perpendicular line to the y axis (a horizontal line), it intersects the graph in many places.  That is, the horizontal line test fails.  Thus the sine function does not have an inverse which is itself a function, at least over the domain [-2π,2π].
Remember that the graph of an inverse relation to a given function is simply the reflection of that graph over the line y = x.  If we do this, the graph shown below results.  The red curve is the function f[x] = Sin[x], the green curve is its inverse, the black line is the line y = x.

From this we see that the domain of the sine function is all real numbers (x ∈ R) and the range is all numbers between -1 and 1 inclusive.  The domain of the inverse relation is all numbers between -1 and 1 inclusive and the range is all real numbers.  

We also see very clearly that the inverse is not a function, however, if we took only a segment of the inverse relation such that the vertical line test would hold, the portions in different colors would be viable choices.  Now to make a good inverse function we want the function to cover all the domain (the range of Sin[x]) and whose range would produce numbers as small as possible.  The only portion then that would do that is the red section centered on the x axis.

So if we defined the sine curve to be only those points between - and , inclusive, and graph the reflection of the sine curve over y = x, we would have the following graph:

The green graph is the graph of the inverse function to y = Sin[x].  This function is called the arcsine and is denoted by y = ArcSin[x] or y = [x].

Facts about

Definition:  The Inverse of Sin[x]
If f[x] = Sin[x] on [,] then  [x] = ArcSin[x] or [x].

The previous sentence is extremely important to understand.  These facts are present:
1)  If the sine function is defined on [,] then the inverse is a functon, otherwise it is only a
     relation.
2)   If the sine function is defined on [,], then the domain of the arcsine function is the set of
     numbers between -1 and 1 inclusive.  The range are all numbers in [,].  Note:  This is why you
      can't ever get an obtuse angle when you hit the keys INV-SIN on your calculator.  The function will
      only return angles between 0 and and between - and 0.
3)  The statements y =  Sin[x] and  x  = ArcSin[y] are equivalent if x ∈ [,].
4)  The statements y =  Sin[x] and  y  = ArcSin[x] are inverses with x ∈ [,].  
     Thus Sin[ArcSin[x]= x  and ArcSin[Sin[x]] = x for x ∈ [,].

The Inverse of Cos[x]

Introduction

As we developed the arcsine function, it wasn't stated exactly what our considerations would be but these things seem to naturally develop:
1)  The inverse graph must be a function,
2)  The inverse graph must be continuous,
3)  The inverse graph must be as close to the x axis as possible.  

We will keep these same considerations in mind as we find the other inverse trig functions.
The graph of y = Cos[x] is shown below in red.  The inverse relation for it is shown in red, green, and blue.  We notice that same difficulty we saw before in that the inverse is not a function.  However, if we divide portions of the curve so that the entire set of numbers between -1 and 1, inclusive is covered, the sections in red, green, and blue result.  Now pick the section closest to the x axis.  Doing this we see that the arccosine relation can be defined to be a function if the range of the inverse is [0,π].  If this is the case then y = Cos[x] must only be defined on [0,π].

Facts about

Definition:  The Inverse of Cos[x]
If f[x] = Cos[x] on [0,π] then  [x] = ArcCos[x] or

1)  If the cosine function is defined on [0,π] then the inverse is a functon, otherwise it is only a relation.
2)   If the cosine function is defined on [0,π], then the domain of the arccosine function is the set of numbers between -1 and 1 inclusive.  The range are all numbers in [0,π].  Note:  This is why you can't ever get a negative angle when you hit the keys INV-COS on your calculator.  The function will only return angles between 0 and π.
3)  The statements y =  Cos[x] and  x  = ArcCos[y] are equivalent if x ∈ [0,π].
4)  The statements y =  Cos[x] and  y  = ArcCos[x] are inverses with x ∈ [0,π].  
     Thus Cos[ArcCos[x]= x  and ArcCos[Cos[x]] = x for all x ∈ [0,π].

The Inverse of Tan[x]

Introduction

Now we will look at the inverse to y = Tan[x].  The graph of y = Tan[x] is shown below in red.  The inverse relation for it is shown in green.  The same difficulties arise as before.  However, if we take only the curve passing through the x axis, we have the ArcTangent defined as a function.  Its domain will be all real numbers and its range will be those numbers in [,].  

Thus the graph of the arctangent function then is:

Facts about

Definition:  The Inverse for Tan[x]
If f[x] = Tan[x] on [,] then [x] = ArcTan[x] or

1)  If the tangent function is defined on [,] then the inverse is a functon, otherwise it is only a relation.
2)   If the tangent function is defined on [,], then the domain of the arctangent function is the set of all real numbers.  The range are all numbers in [,].  
3)  The statements y =  Tan[x] and  x  = ArcTan[y] are equivalent if x ∈ [,].
4)  The statements y =  Tan[x] and  y  = ArcTan[x] are inverses with x ∈ [,].  
     Thus Tan[ArcTan[x]= x  and ArcTan[Tan[x]] = x for all x ∈ [,]

Brief Summary of All Three

The remaining inverse functions can all be defined in a similar manner, however, since their use is limited and one can always convert one of the first three, they will not be included in this discussion.   Therefore we have these inverse trig functions:

Remember these facts:

1)  y = ArcSin[x] ⇔ x = Sin[y]  where y ∈ [-,].
2)  y = ArcCos[x] ⇔ x = Cos[y]  where y ∈ [0,π].
3)  y = ArcTan[x] ⇔ x = Tan[y]  where y ∈ [-,].

Operations using the Inverse Trig Functions

Graphing these functions

Graphing these functions is done like you have graphed any other function.  

Example 1:  Sketch the graph of y = 4 + 2 ArcSin[3(x-1)]
Solution:  
Method 1:  Since the domain of the arcsine is [-1,1] then -1 ≤ 3(x -1) ≤ 1  ⇒ ≤ x - 1 ≤   => ≤ x ≤ .  
Now we use these points to generate a graph.  

Method 2:  We could also do this by stretching, shrinking, and shifting the graph of y = ArcSin[x]  (shown in red below).  
y = 4 + 2 ArcSin[3(x-1)] is the same as y - 4 = 2 ArcSin[3(x-1)].   Now the graph of y = ArcSin[3x] would cause all x coordinates to be multiplied by , thus the graph is shrunk horizontally  (green).  If we have y = 2 ArcSin[3x] then all y coordinates would be multiplied by 2, causing a vertical stretch by a factor of 2 (blue).  Replacing x with x -1, causes a horizontal shift of 1 unit to the right (grey) and then finally replacing y with y - 4, moves the graph 4 units up (black).  The graphs shown below give this transition.

Simplifying Expressions

Now having the inverse functions at our disposal we can simplify expressions and solve trig equations.  

Example 2:  Find ArcSin[-]

Solution:  This asks us "What angle has a sine of -?"  This is a 60 degree angle or radians in the fourth quadrant.  Since the ArcSin function is only defined for values in [-,], we must answer this with -.

Example 3:  Simplify  Sin[ArcCos[]]
Solution:  This statement asks "What is the sine of the angle which has a cosine of ?"  A sketch will help a great deal.  Since the cosine fo the angle is positive we now that the angle must be in the first or fourth quadrants.  Since the arccosine is defined for 1st and 2nd quadrant angles only, then the angle we want is in the first quadrant.  Thus we can draw our triangle there.

Using the Pythagorean theorem we find y = 12.  Therefore the sine of the angle is .  
So Sin[ArcCos[]] = .

Example 4:  Simplify  Tan[ArcSin[x]]
Solution:  This statement wants to know what the tangent of the angle is, if the sine of the angle is x.  Again we sketch a triangle.

Since the sine of the angle is x we have the opposite side to be x and the hypotenuse to be 1.  Using the Pythagorean theorem to find the adjacent side, we have it to be .  Thus the tangent of the angle is .  
Simplifying this we have Tan[ArcSin[x]] = .

Example 5:  Find [ Sin[-] ]

Solution:  This asks "What is angle whose cosine is the sine of -?"  We first begin by finding the sine of -.  This is a 30 degree angle in the fourth quadrant, thus the sine is -.  Now we find ArcCos[-].  This is a 60 degree angle in the 2nd quadrant, so [ Sin[-] ] =

Using the inverse functions to solve equations

Example 6:  Solve this equation for x.    y = 4 + 2 Sin[4(x -1)]
Solution:    y - 4 = 2 Sin[4(x -1)]
              = Sin[ 4(x -1) ]
              [ ] = [Sin[ 4(x-1)] ]
              [ ] = 4(x-1)
              [ ] = x - 1
              1 + [ ] = x

Assignment

Part I.  Find the exact value of each expression and give the results in terms of π.

  1) ArcCos[-]               2)  []                   3) [-]

Part II.  Find the value of each of the following, in radians, to the nearest thousandths.

  4)  ArcSin[-2/3]                  5)  [.8]                6) ArcTan[6]

Part III. Evaluate each of the following. All angles are and need to be expressed in radians.

  7) Sin[ArcCos[]]            8)  Cos[[-.25]]           9) [Cos[-.25]]
  10) [Sin[0.56]]            11)  Sec[ArcCos[-]]        12) Tan[[] ]

Part IV.  Find an algebraic expression for each of the following

13) Tan[ArcSin[x]]              14)  Sin[[x]]         15)  Csc[ArcSin[x]]

Part V.  Graph the following with the domain in radians.

16)  f[x] = 5 +  4 [0.5(x+3)]       17)  f[x] = -1 + 2 ArcSin[ (x - ) ]

Part VI.  Answer the questions posed below.

18)  Scientists sometimes use a formula like T[t] = a Sin[bt + c] + d to approximate the temperature during the day.  Suppose T[t] = 18 Sin[t - ) + 66 gives the temperature at location A, where t is measured in hours and t = 0 corresponds to midnight. What is the highest temperature location A can acheive?  What is the lowest temperature?  When will both of those temperatures occur?  How hot will it be at 5pm?  When will it be 66 degrees?

Created by Mathematica  (September 20, 2003)