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Module 5-II – Trigonometric Functions

5-II.4 – Inverse Trig Functions, Equations and Inequalities

 

Introduction

 

Both in module 5-I and in section 5-II.1 of this module we have briefly addressed the problem of finding angles from given trig ratios. This amounts to solving for t in an equation like  where v is a given value. In this section we shall look at such problems in more detail and will also find angles from more complicated equations involving their trig ratios.

           

Section Learning Objectives

After completing this section you should be able to:

·        Solve simple trig equations either by recognizing particular special cases and/or by reference to “inverse trig functions”

·        Use identities and algebraic manipulation to solve more complicated equations

·        Solve various applied problems involving trigonometric equations and inequalities

           

What to Do

 

1.         Read section 6.2 of the text.

2.         Read the following Study Notes.

3.         Read section 6.6 of the text.

4.         Read the Follow-up Notes and Discussion

5.         Follow the instructions regarding Further Practice.

 

 

 

Study Notes

 

In the text section 6.2 (that you have just read), the basic equations like  (and similarly with cos and tan) were combined with algebraic steps like factoring to solve more complicated equations. (This is much like what we did in Unit 4.5 with exponential and logarithmic equations.) But everything still depends on the solution of those basic equations and we only have exact solutions for special cases.

 

To solve  for general v we have been relying on the calculator’s  button to get an approximate solution, so it would be a good idea to look more closely at what this button actually does.(Especially since we know that trig functions all fail the HLT and so don’t actually have inverse functions at all – so the name on that button must be a lie! )

 

We know that when the calculator is in radian mode the  button applied to input v produces a solution of the equation . But we also know that there are infinitely many possible solutions. Clearly the button picks just one – but which one?

 

The answer, as discussed in more detail in the text’s section 6.6, is that the result produced by the  button is always between  and ,(i.e. it always corresponds to a positive or negative acute angle). What does this correspond to in terms of the circle picture? (see answer #1)

 

Although the sine function is not invertible, the function  defined by restricting the sine function to the interval , is one-to-one and so does have an inverse function. Its graph consists of just that part of the sine graph between  and . This does pass the HLT, and its reflection across the diagonal gives the graph of its inverse function. Can you draw these? (see answer #2)

 

The function defined by the calculator’s  button is just exactly the inverse function  of the above defined function S. It is not really the inverse function of the sine itself (in fact there isn’t one), but nonetheless it is often denoted by  and called the “inverse sine function”. Since this is not strictly correct, many authors prefer a different name and call it the arcsine function as discussed on p511 of the text.

 

The corresponding stories for the calculator’s  and  buttons are similar, except that tan is undefined at  and , so the restriction is to just the open interval , and for cos, the interval between  and  does not work, and so  is used instead. Try reflecting the graph of  across the diagonal, and compare the result with you got for the arcsine above. (see answer #3)

Doing the same thing for the arctan leads to the graph in Figure 8 on page 515 of the text.

 

 

 

Follow-up Notes and Discussion

 

In your reading of section 6.6 of the text, you will have seen that            .

The restriction is necessary because the left hand side (LHS) is undefined unless x is in the domain of the arcsine (which is the same as the range of the sine).

 

You will also have seen that

            .

In this case though, the LHS is defined for all x, (since  is defined for all x and is always between -1 and 1 and so in the domain of  ). But if , then , which is in that interval, cannot be the same as x, which is not.

 

Similar comments apply to the compositions of cos with arccos, and tan with arctan.

 

Try to sketch the graphs of  and  without a calculator

(then use a calculator or the GraphExplorer to check your answer).

 

The text’s Examples  4, 6, and 8 all involve composition of one trig function with the inverse of another. This is a fancy way of asking you to find other trig ratios of an angle for which one ratio is given (as you did way back near the beginning of our trig study in Unit 5.3). Pay attention to the identity proved in Example 8. That is something you will find important later on when studying Calculus.

 

 

Further Practice

 

Check your understanding, and practice for speed, by working through some of the Exercises on pp511-514 and 469-472.

 

Do enough of the odd numbered questions of each type to convince yourself that you can get the right answers. (Note that, as usual, the answers are in the back of the text and complete worked solutions are in the student study guide - but try to avoid looking at answers or solutions until you have made your own best effort)

 

As a minimum, you should try the following: From section 6.2 #1,5,31,35,55, and 75,

and from section 6.6 #1,11,21,31,41,51,61,67, and 71.