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Graphing Inverse Trigonometric Functions

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Graphs of Inverse Trigonometric Functions

In order for inverses of functions to be functions, the original function must pass the horizontal line test.  Though none of the trigonometric functions pass the horizontal line test, you can restrict their domains so that they can pass.  Then the inverses are produced just like with normal functions.  Once you have the basic inverse functions, the normal transformation rules apply. 

Why is \sin^{-1} (\sin 370^\circ) \neq 370^\circ ?  Don’t the arcsin and sin just cancel out? 

Guidance

Since none of the six trigonometric functions pass the horizontal line test, you must restrict their domains before finding inverses of these functions.  This is just like the way y=\sqrt{x}  is the inverse of y=x^2  when you restrict the domain to x \ge 0 .

Consider the sine graph:

As a general rule, the restrictions to the domain are either the interval \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]  or [0, \pi]  to keep things simple.  In this case sine is restricted to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] , as shown above.  To find the inverse, reflect the bold portion across the line y=x . The blue curve below shows f(x)=\sin^{-1}x .

The result of this inversion is that arcsine will only ever produce angles between -\frac{\pi}{2}  and \frac{\pi}{2} .  You must use logic and common sense to interpret these numbers in context.

Example A

What is the graph of f(x)=\cos^{-1}x ?

Solution:  Graph the portion of cosine that fits the horizontal line test (the interval [0, \pi] ) and reflect across the line y=x .

Example B

Graph the function f(x)=2 \cos^{-1} (x-1) .

Solution:  Since the graph of f(x)=\cos^{-1}x  was done in Example A, now you just need to shift it right one unit and stretch it vertically by a factor of 2.  It intersected the  x axis at 1 before and now it will intersect at 2.  It reached a height of \pi  before and now it will reach a height of 2 \pi .

Example C

Evaluate the following expression with and without a calculator using right triangles and your knowledge of inverse trigonometric functions.

\cot \left(\csc^{-1} \left(-\frac{13}{5}\right)\right)

Solution:  When using a calculator it can be extremely confusing trying to tell the difference between \sin^{-1}x  and (\sin x)^{-1} .  In order to be able to effectively calculate this out it is best to write the expression explicitly only in terms of functions that your calculator does have. 

The hardest part of this question is seeing the csc as a function (which produces an angle) on a ratio of a hypotenuse of 13 and an opposite side of -5.  The sine of the inverse ratio must produce the same angle, so you can substitute it.

  • \csc^{-1} \left(-\frac{13}{5}\right)=\sin^{-1} \left(-\frac{5}{13}\right)
  • \cot (\theta)=\frac{1}{\tan \theta}

\cot \left(\csc^{-1} \left (-\frac{13}{5}\right) \right)=\frac{1}{\tan \left(\sin^{-1} \left(-\frac{5}{13}\right)\right)}=-\frac{12}{5}

Not using a calculator is usually significantly easier.  Start with your knowledge that \csc^{-1} \left(-\frac{13}{5}\right)  describes an angle in the fourth or the second quadrant because those are the two quadrants where cosecant is negative.  Since \csc^{-1} \theta  has range -\frac{\pi}{2}, \frac{\pi}{2} , it only produces angles in quadrant I or quadrant IV (see Guided Practice 2).  This triangle must then be in the fourth quadrant.  All you need to do is draw the triangle and identify the cotangent ratio.

Cotangent is adjacent over opposite.

\cot \left(\csc^{-1} \left(-\frac{13}{5}\right)\right)=-\frac{12}{5}

Concept Problem Revisited

Since arcsine only produces angles between -\frac{\pi}{2}  and \frac{\pi}{2}  or -90^\circ  to +90^\circ  the result of \sin^{-1} (\sin 370^\circ)  is 10^\circ  which is coterminal to 370^\circ

Vocabulary

Restricted domain refers to the fact that when creating an inverse you sometimes must cut off the domain of most of the function, saving the largest possible portion so that when the inverse is created it is also a function. 

Guided Practice

1. What is the graph of y=\tan^{-1} x ?

2. What is the graph of y=\csc^{-1}x ?

3. Evaluate the expression \csc \left(\cot^{-1} \left[-\frac{8}{6}\right]\right) .

Answers:

1. Graph the portion of tangent that fits the horizontal line test and reflect across the line y=x .  Note that the graph of arctan is in blue.

2. Graph the portion of cosecant that fits the horizontal line test and reflect across the line y=x .

Note that  f(x)=\csc^{-1}x is in blue.  Also note that the word “arc-co-secant” is too cumbersome to use because of the train of prefixes. 

3.

\csc \left(\cot^{-1} \left[-\frac{8}{6}\right]\right)&=\frac{1}{\sin \left(\cot^{-1} \left[-\frac{8}{6}\right]\right)}\\&= \frac{1}{\sin \left(\tan^{-1} \left(-\frac{6}{8}\right)\right)}\\&= -\frac{10}{6}\\&= -\frac{5}{3}

Practice

1. Graph f(x)=\cot^{-1}x .

2. Graph g(x)=\sec^{-1}x .

Name each of the following graphs.

3.

4.

5.

6.

7.

Graph each of the following functions using your knowledge of function transformations.

8.  h(x)=3 \sin^{-1} (x+1)

9.  k(x)=2 \sin^{-1}(x)+\frac{\pi}{2}

10.  m(x)=-\cos^{-1}(x-2)

11.  j(x)=\cot^{-1}(x)+\pi

12.  p(x)=-2 \tan^{-1}(x-1)

13.  q(x)=\csc^{-1}(x-2)

14. r(x)=-\sec^{-1}(x)+4

15. t(x)=\csc^{-1}(x+1) -\frac{3 \pi}{2}

16. v(x)=2 \sec^{-1}(x+2)+\frac{\pi}{2}

17. w(x)=-\cot^{-1}(x)-\frac{\pi}{2}

Evaluate each expression.

18. \sec \left(\tan^{-1} \left[\frac{3}{4}\right]\right)

19. \cot \left(\csc^{-1} \left[\frac{13}{12}\right]\right)

20.  \csc \left(\tan^{-1} \left[\frac{4}{3}\right]\right)

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