Solving Trig Equations with Calculators, Part I : The single problem along with the equations we solved out in there is that they pretty much all had solutions which came from a handful of "standard" angles and certainly there are several equations out there that simply don't.  Hence, in this section we will look at some more trig equations, the majority of which will need the use of a calculator to solve (a couple won't necessitate for calculator).

The issue that we are using calculators in this section does not though mean that the problems in the earlier section aren't important.  This is going to be supposed in this section that the fundamental ideas of solving trig equations are known & that we don't have to go back over them here. In specific, it is supposed that you can utilizes a unit circle to help you determine all answers to the equation (although the procedure here is a bit different as we'll illustrates) and it is supposed that you can determine answers in a given interval.

Before going through with the problems we have to go over how our calculators work so that we can get the accurate answers.  Calculators are great tools although if you don't know how they work &how to interpret their answers you can put in serious trouble.

Firstly, as already pointed out in earlier sections, everything we are going to be doing in this will be in radians so ensure that your calculator is set to radians before trying the problems in this section.  Also, here we are going to use four decimal places of accuracy in the work.  You can utilize more if you desire, but we'll always use at least four decimal places of accuracy.

Next, and somewhat more significantly, we have to understand how calculators calculate answers to inverse trig functions. We didn't cover inverse trig functions, although they are just inverse functions & we have talked a little regarding inverse functions in a review section. The only actual difference is that now we are using trig functions. We'll simply be looking at three of them and they are following:

Inverse Cosine: cos^-1( x )=arccos( x )

Inverse Sine    : sin ^-1( x)=arcsin( x )

Inverse Tangent: tan ^-1 ( x )=arctan( x )

As illustrated there are two different notations which are commonly utilized.  In these notes we'll be using the first form as it is little more compact.  Most of the calculators these days will have buttons on them for these three hence ensure that yours does as well.

Now we need to deal with how calculators give answers to these. Let's imagine, for instance, that we wanted our calculator to compute cos^-1(  3/4) .  Firstly, remember that what the calculator is in fact computing is the angle, let's say x, that we would plug into cosine to obtain a value of    3 /4, or

x = cos^-1( 3/4) ⇒cos( x )= 3/4

Hence, in other words, while we are using our calculator to compute an inverse trig function we are actually solving a simple trig equation.

Having our calculator calculate cos^-1(  3) and therefore solve cos( x )=¾  gives,

x = cos^-1( 3)= 0.7227

From the earlier section we know that there have to in fact be an infinite number of answers to this including a second angle that is in the interval [0, 2π] .  However, our calculator just gave us a single answer.  How to find out what the other angles are will be covered in the given examples hence we won't go into detail here regarding that. We did have to point out though, that the calculators will just give single answer & that we're going to contain more work to do than just plugging a number into calculator.

As we know that there are assumed to be an infinite number of solutions to cos( x)=¾ the next question we have to ask then is just how did the calculator decide to return the answer that it did? Why this one & not one of the others?  Will this give the similar answer every time?

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

0 ≤ cos_-1( x) ≤ π- π/2 ≤ sin ^-1(x )≤ π/2- π/2-1(x )<π/2

If you think back of the unit circle and remember again that we think of cosine as the horizontal axis then we can illustrates which we'll cover all possible values of cosine in the upper half of the circle and it is exactly the range given above for the inverse cosine.  Similarly, as we think of sine as the vertical axis in the unit circle we can illustrates that we'll cover all probable values of sine in the right half of the unit circle & that is the range given above.

For the tangent range consider the graph of the tangent function itself and we'll see that one branch of the tangent is covered into the range given above & so that is the range we'll use for inverse tangent.  Note that we don't involve the endpoints in the range for inverse tangent as tangent does not exist there.

Thus, if we can remember these rules we will be capable to determine the remaining angle in [0, 2π]that also works for each solution.