We’re now going to complete our study of inequalities by considering inequalities that involve trigonometric terms.
They’re trickier, it must be said, than exponential or logarithmic inequalities, in general. In fact, I would say they’re pretty much at the cutting edge of what people who have taken their pre-calculus course can do. One of the reasons they’re harder is because monotonicity, like we had for exponential functions, is far less useful than it was. And it’s replaced by periodicity. And also, of course, as usual in trigonometry, you need to have a good grasp of all those identities and properties that we know hold for trigonometric functions. As a general principle, one can consider the inequality first on a suitable interval covering the period of the functions involved and then find the other solutions of the inequality by invoking periodicity.
These possibly vague remarks should become clearer as we look at a certain number of examples. Let’s begin with a simple inequality of the type cosine x greater or equal alpha, or possibly less than or equal alpha. The first remark is that if the absolute value of alpha is bigger than 1, then the solution set will be obvious. It’ll be either the real line or the empty set. Why? Well, because cosine x greater than minus 2, for instance, is always true for any x. Or cosine x greater than 3, for example, is impossible. So the interesting case, really, is when the absolute value of alpha is less than or equal 1.
In that case, there’s a particular angle a whose cosine is equal to alpha. It’s the angle we called arccosine alpha, and it lies, as you recall, in the interval 0 pi. So the two inequalities we’re looking at, then, will be cosine x greater or equal to cos a or less than or equal. Let’s start with the first. How do we solve this inequality? One way to do it is to look at the graph of the cosine function, which is something we should know very well, indeed. Now here’s the angle a. It’s between 0 and pi, and it generates a certain value of cosine.
And we’re looking for those points for which the cosine of x will be at least equal to that same value. Well, if you look around the point a, you can see that those points will be to the left of a until you reach a certain other angle. What is that other angle? Well, you can see by symmetry that it’s the angle minus a, the one that has the same cosine as a.
Also, you can see that in the interval minus pi to pi, which is an interval of length 2 pi, which is the period of the function involved, the only solutions for which cos x will be greater or equal to cos a are those in the interval minus a to a. And that tells you where x has to lie in order to be a solution of our inequality relative to the interval minus pi to pi. Of course, given that cosine is periodic of period 2 pi, there’ll be lots of other x’s satisfying the same inequality.
They’ll be in some other interval, and we can capture all of those by simply adding to the ones we’ve already identified any integral multiple of 2 pi, where the integer k is either positive or negative. Notational convention here, when I take an interval minus -a, a, and I add a number like 2 k pi, that means I’ve taken the interval consisting of all numbers in the original interval to which I’ve added that. Another way of saying it– it’s the interval from minus a plus 2 k pi to a plus 2 k pi. Notational convention. Now let’s look at the other case, cosine x less than or equal cos a.
It can be solved in rather similar fashion by looking at the graph of the cosine function. Here’s the angle a. It generates a certain cosine. Now we’re looking for the x’s whose value of cosine is less than cosine a. And we see that off on the right, off to a certain angle, which we see by symmetry is 2 pi minus a, those cosines will be less than or equal than what they were at a. And we further see that relative to the interval 0, 2 pi, these are the only x’s, the ones in that interval, a to 2 pi minus a, for which the cosine will be less than it was at a.
This immediately gives us a first estimate of our set of solutions, the interval a to 2 pi minus a. But, of course, there will be many other solutions in other intervals of length 2 pi. And we capture those by adding to that interval 2 k pi, where k is any positive or negative integer. So we have found the solutions in both cases of our simple inequality. Let’s use an example of how we can apply that to the inequalities such as cos x less than minus 1/2. What do we have to do first? Well, we have to find an angle whose cosine is minus 1/2. Easy enough to do. It’s 2 pi over 3.
So we’re solving cos x less than cosine 2 pi over 3. We could do it graphically, as I’ve just illustrated. But there’s another technique I’d like to show you. It uses the circular definition of sine and cosine, the unit circle thing. If you look at the angle 2 pi over 3, and you look at the point that corresponds to it on the unit circle, and the cosine minus 1/2 is the x-coordinate of that point. What are the points on the unit circle which will have an x-coordinate less than minus 1/2? Well, you can identify those fairly easily. They’ll be the ones to the left of that vertical segment.
And the blue point that gives you the bottom point there corresponds to the angle minus 2 pi over 3. Remember, cosine is even, and the cosine of minus 2 pi over 3 will be the same as the cosine of 2 pi over 3. Now what points are you trying to capture? Those points that lie to the left of that vertical segment. How do you express that vertical segment? What angles are in it? Well, it’ll be the angles between this one. I’ve now expressed my angle minus 2 pi over 3 in an equivalent form as 4 pi over 3. It will be the angles up to there but starting at 2 pi over 3.
In other words, it’ll be all angles between 2 pi over 3 and 4 pi over 3. And that’s our solution set. Except that periodicity will give us other solutions. So, of course, I need to add to that any integral multiple of 2 pi, positive or negative. We could have solved that by looking at the graph of cosine, also, the alternate method. Some people find it easier. Some people prefer the unit circle. How would it have worked graphically? Well, we look at our angle 2 pi over 3. It generates a certain cosine value. We’re looking for x’s that give you something less than that. They’re obviously off on the right. They go all the way to 4 pi over 3.
And when we look at the interval 0, 2 pi, we see that those are the only x’s that will generate a cosine less than minus 1/2. And that gives us the same answer. Now, exactly similar reasoning as we’ve applied to cosine we’ll apply to sine. And we can solve inequalities such as sine x less than or equal sine a much the same way. Now the a will be between minus pi over 2 and pi over 2, because that’s where the arcsine function takes its values. A simple example. Find the x’s for which sine x is less than or equal root 3 over 2. Well, we begin by finding an angle whose sine is root 3 over 2.
That’s, of course, pi over 3. It’s one of our known values that we know by heart. And how do we proceed? Well, graphically, we look at pi over 3, and it generates a certain sine. And we’re looking for the x’s for which the sine is no bigger. So clearly that has us moved to the left all the way to a certain angle, which has the same sine as pi over 3. What is that angle? It’s the supplementary angle. In this case, minus 4 pi over 3. Now, in the interval minus pi to pi, it’s not quite clear that we have the only solutions.
So it’s better to shift the interval over to the left so that we obviously have captured all the possible solutions of an inequality in an interval of length 2 pi. That interval, of course, is the interval minus 3 pi over 2 to pi over 2. Now we can be quite confident that in that interval, the only solutions are the ones in the designated interval. So we take that interval, but, of course, we then proceed to add 2 k pi to capture by periodicity all the other solutions.
We could have solved this problem by the unit circle approach. Let’s indicate how it would work. This is the picture you would be using to solve this inequality. What does that picture correspond to? Well, first, we identified the angle pi over 3, and we placed a point on the unit circle to which it corresponds. Then we found the angle for which the sine was the same, namely the supplementary angle 2 pi over 3. And now we look at the points that we want to identify as solving our inequality on the unit circle. That means that the y-coordinate, the sine of x, should be no bigger than it is for those two points.
That means that we’re not in the sector up above. We’re rather in the complement of that sector. How do we express the complement of that sector? Well, we see that it consists of all angles between minus 4 pi over 3 and pi over 3, which gives us that interval as our initial set of solutions. And then we apply periodicity. We add 2 k pi to that to find the set of all solutions. And, of course, we get the same answer as we did by the graphical method. Now you notice that even in solving these simple inequalities, you’ve had to really take account of the method involved.
And you’ve had to use quite a bit of thought all the way through, either graphically or with unit circles. I would say that if you can solve these kinds of inequalities correctly, you’re in the inequality elite.