Solving Trig Equations with Calculators, Part I

Review : Solving Trig Equations with Calculators, Part I

In the previous section we started solving trig equations. The only problem with the equations we solved in there is that they pretty much all had solutions that came from a handful of �standard� angles and of course there are many equations out there that simply don�t. So, in this section we are going to take a look at some more trig equations, the majority of which will require the use of a calculator to solve (a couple won�t need a calculator).

The fact that we are using calculators in this section does not however mean that the problems in the previous section aren�t important. It is going to be assumed in this section that the basic ideas of solving trig equations are known and that we don�t need to go back over them here. In particular, it is assumed that you can use a unit circle to help you find all answers to the equation (although the process here is a little different as we�ll see) and it is assumed that you can find answers in a given interval. If you are unfamiliar with these ideas you should first go to the previous section and go over those problems.

Before proceeding with the problems we need to go over how our calculators work so that we can get the correct answers. Calculators are great tools but if you don�t know how they work and how to interpret their answers you can get in serious trouble.

First, as already pointed out in previous sections, everything we are going to be doing here will be in radians so make sure that your calculator is set to radians before attempting the problems in this section. Also, we are going to use 4 decimal places of accuracy in the work here. You can use more if you want, but in this class we�ll always use at least 4 decimal places of accuracy.

Next, and somewhat more importantly, we need to understand how calculators give answers to inverse trig functions. We didn�t cover inverse trig functions in this review, but they are just inverse functions and we have talked a little bit about inverse functions in a review section. The only real difference is that we are now using trig functions. We�ll only be looking at three of them and they are:

As shown there are two different notations that are commonly used. In these notes we�ll be using the first form since it is a little more compact. Most calculators these days will have buttons on them for these three so make sure that yours does as well.

We now need to deal with how calculators give answers to these. Let�s suppose, for example, that we wanted our calculator to compute . First, remember that what the calculator is actually computing is the angle, let�s say x, that we would plug into cosine to get a value of , or

So, in other words, when we are using our calculator to compute an inverse trig function we are really solving a simple trig equation.

Having our calculator compute and hence solve gives,

From the previous section we know that there should in fact be an infinite number of answers to this including a second angle that is in the interval . However, our calculator only gave us a single answer. How to determine what the other angles are will be covered in the following examples so we won�t go into detail here about that. We did need to point out however, that the calculators will only give a single answer and that we�re going to have more work to do than just plugging a number into a calculator.

Since we know that there are supposed to be an infinite number of solutions to the next question we should ask then is just how did the calculator decide to return the answer that it did? Why this one and not one of the others? Will it give the same answer every time?

There are rules that determine just what answer the calculator gives. All calculators will give answers in the following ranges.

If you think back to the unit circle and recall that we think of cosine as the horizontal axis the we can see that we�ll cover all possible values of cosine in the upper half of the circle and this is exactly the range give above for the inverse cosine. Likewise, since we think of sine as the vertical axis in the unit circle we can see that we�ll cover all possible values of sine in the right half of the unit circle and that is the range given above.

For the tangent range look back to the graph of the tangent function itself and we�ll see that one branch of the tangent is covered in the range given above and so that is the range we�ll use for inverse tangent. Note as well that we don�t include the endpoints in the range for inverse tangent since tangent does not exist there.

So, if we can remember these rules we will be able to determine the remaining angle in that also works for each solution.

As a final quick topic let�s note that it will, on occasion, be useful to remember the decimal representations of some basic angles. So here they are,

Using these we can quickly see that must be in the first quadrant since 0.7227 is between 0 and 1.5708. This will be of great help when we go to determine the remaining angles

So, once again, we can�t stress enough that calculators are great tools that can be of tremendous help to us, but it you don�t understand how they work you will often get the answers to problems wrong.

So, with all that out of the way let�s take a look at our first problem.

Example Solve on[-8,10].

Solution

Okay, the first step here is identical to the problems in the previous section. We first need to isolate the cosine on one side by itself and then use our calculator to get the first answer.

So, this is the one we were using above in the opening discussion of this section. At the time we mentioned that there were infinite number of answers and that we�d be seeing how to find them later. Well that time is now.

First, let�s take a quick look at a unit circle for this example.

The angle that we�ve found is shown on the circle as well as the other angle that we know should also be an answer. Finding this angle here is just as easy as in the previous section. Since the line segment in the first quadrant forms an angle of 0.7227 radians with the positive x-axis then so does the line segment in the fourth quadrant. This means that we can use either -0.7227 as the second angle or . Which you use depends on which you prefer. We�ll pretty much always use the positive angle to avoid the possibility that we�ll lose the minus sign.

So, all possible solutions, ignoring the interval for a second, are then,

Now, all we need to do is plug in values of n to determine the angle that are actually in the interval. Here�s the work for that.

So, the solutions to this equation, in the given interval, are,