In section 1.4, the topic is trig functions and their inverses. So in calculus, we often use right triangle trigonometry. So as far as right triangle trig goes, to make a long story short, right triangles possess ratios which depend only on their central angle. So I'm going to call this angle theta here the central angle.
And if I were to stretch or shrink this triangle and keep this angle here the same, There are a common set of ratios that we talk about that will stay the same. And so these ratios depend only on this angle. They do not depend on the size of the triangle.
And so these ratios have their own names. So those names are the sine, cosine, and tangent. And then the cosecant, the secant, and the cotangent.
So those are defined this way. The sine of theta is defined as the opposite. divided by the hypotenuse.
The cosine of theta is defined as the adjacent over the hypotenuse, and the tangent of theta is defined as the opposite divided by the adjacent. And what we also see is that the tangent is the ratio of the sine to the cosine. We also have what we call the reciprocal ratios or the reciprocal trig functions.
Those are the cosecant, which is the reciprocal of sine, the secant, which is the reciprocal of cosine, and then the cotangent, which is the reciprocal of the tangent. Be careful not to confuse reciprocal with inverse. Those are two different ideas. We'll talk about inverse trig functions later on in this video. So let's go ahead and write the six trig ratios for the following right triangle.
So very simply, the sine of theta is going to be the opposite divided by the hypotenuse. That's going to be 4 divided by 5. The cosine of theta is going to be the adjacent, 3, divided by the hypotenuse, 5. And then the tangent of theta is going to be the opposite divided by the adjacent. So that's going to be 4 divided by 3. And now very easily we can write the reciprocal ratios. The cosecant is the reciprocal of sine, so we have 5 over 4. The secant is the reciprocal of cosine, so we have 5 over 3. And then the cotangent is the reciprocal of tangent, we have 3 over 4. So for this triangle, here are our six trig ratios.
So as a student in Math 1060, you need to know the unit circle. The unit circle gives us the values of sine and cosine and consequently the tangent for some common angles. And so what the unit circle does is it inscribes a triangle whose hypotenuse is equal to one All right, and so this triangle here, in calculus we use, typically we use radians for our degree measure.
So this triangle here has a central angle of 30 degrees, or pi over 6 radians, and so the unit circle shows us that the cosine, the cosine is the x-coordinate, so the cosine is this value on the bottom, and then the sine is the y-coordinate, the value on the side here. And so the unit circle gives us the values of sine and cosine for some common angles that we often see. So you need to know the unit circle by heart.
So again, the x-coordinate is your cosine. The y-coordinate is your sine. And then since the tangent is the sine over the cosine, y divided by x will give you the tangent. So I find it easiest just to memorize one quadrant. Notice that all the numbers are very similar.
The only thing that changes is the sine attached based on the quadrant. So in quadrant 2, cosine is negative. In quadrant 3, they're both negative.
And in quadrant 4, the sine is negative. So for me, it's easiest just to remember one quadrant here and adjust everything else based on sine. So you need to have the unit circle memorized by heart. You need to be able to quickly produce the value of sine and cosine for common angles.
And these common angles are given here in the unit circle. All right, so... Let's find, let's apply some ideas here to find the cosine and tangent of some angle theta given that the sine of theta is 3 over 5 and that the angle theta is in the interval pi over 2 to pi. So the interval pi over 2 to pi is going to be the second quadrant. So that's going to tell us something about how we're going to write, how we're going to attach a sine.
SIG into some of these ratios. So first off let's go ahead and draw a triangle. So drawing a right triangle, we'll write the angle theta and so we're told that the sine of theta is 3 over 5. So we know that the sine is the opposite over the hypotenuse. So we know that 3 and 5 go here and we have a missing side to find. So applying the Pythagorean theorem, we're going to have 3 squared plus x squared equals 5 squared.
That's going to be 9 plus x squared equals 25. Subtracting 9, that means x squared has to be 16. And so x is going to be plus and minus 4, but for the sake of my triangle here, I'm just going to take the positive 4 and we will adjust the sine based on the quadrant. So now I can go ahead and write the cosine of theta. The cosine is going to be the adjacent over the hypotenuse.
So we're going to have 4 over 5. But we're in quadrant 2. And in quadrant 2, the cosine, our x-coordinate, is negative. So I need to attach a negative to this 4 over 5. And now we can find the tangent of theta. The tangent of theta is going to be the opposite over the adjacent. So 3 divided by 4. Now we are again in the second quadrant.
In the second quadrant, sine is positive, cosine is negative. The tangent is the ratio of sine to cosine, so a positive divided by a negative is also going to be negative. So we want to adjust the sine here based on the quadrant that we were given that this angle was lying in, and that was quadrant 2. All right, some trig identities. In Math 1060, you typically don't need to memorize them.
If you need them as a part of an exam or an assignment, we will typically give them to you. But be aware of some of the identities and perhaps where to reference them. Perhaps the most famous identity here is the Pythagorean identity sine squared of theta plus cosine squared of theta equals one.
You can verify that by looking at the unit circle. And also these other two are a consequence of the first. The second one here results in dividing the first by cosine squared.
And then this the last one here results in dividing the first by sine squared. So these two are a consequence of the first. These are the Pythagorean identities. There's also the double angle identities.
and the half angle identities. Again, you don't need to memorize them, but be aware of these identities. So some facts about the trig functions.
So a function is called periodic if there exists some number, and that number not equal to 0 so that we have this relationship. So f of x plus p is equal to f of x for all x in the domain. So in other words, there's some point, there's some number that when I add that number to this value x, the value is unchanged. The output is unchanged. And that number is called the period.
So by definition here, a periodic function is not one-to-one. And in fact, well, a periodic function is never one-to-one. Because notice here that I have two different inputs giving me the same output.
So a periodic function is never one-to-one. Here's the domain range in period of the sine, cosine, and tangent functions. Probably the most important here is to know the domain range and period of the sine and the cosine. The tangent function gets a little bit weird because the cosine function takes on values of 0. And since the tangent function is a ratio, 0 being in the bottom of our ratio presents some problems.
So perhaps most importantly, sine and cosine know their domain. range and period and then we have one thing also here the tangent function has vertical asymptotes due to that denominator being 0 occasionally and so that happens at these values pi over 2 3 pi over 2 5 pi over 2 and so on. Here's the domain range and period for the reciprocals.
So these start to get kind of ugly and kind of difficult to memorize. So know where to reference these but I don't think it's necessary to have these particular, this particular information about the reciprocal trig functions memorized. So let's solve some trig equations. So let's look at this equation sine squared of theta plus two sine of theta plus two equal to 1. And we're just going to restrict ourselves to the interval 0 to 2 pi. So how do we solve this equation?
Notice that this equation has some sort of quadratic flavor to it. So I often, to make this a little bit more obvious, let's change something here. So let's let x equal the sine of theta. So our equation sine squared of theta plus 2 sine of theta plus 2 equals 1 is going to be converted to this is going to be converted to x squared plus 2x plus 2 equals 1. Alright, so how do we solve a quadratic?
We want to set it equal to zero, so I'm going to subtract the 1 over. I get x squared plus 2x plus 2 minus 1 is equal to 0. And this is going to factor. This is going to factor into x plus 1 and x plus 1. So notice here that we have a repeated factor.
So we could also write this as just x plus 1 squared equals 0. So how do I solve this? Well, I want to set one of those factors equal to 0. So I have x plus 1 equals 0, but x was equal to the sine of theta. All right, so let's replace that sine of theta plus 1 equals 0 and subtract the 1 over, I have this expression sine theta equals negative 1. So I want to think about which angle has a sine of negative 1 on the interval 0 to 2 pi. And this is where you need to know the unit circle. The sine is the y-coordinate.
And where is the y-coordinate negative 1? That's going to happen at the angle 3 pi over 2. So let's look at another trig equation. Here I have tangent squared of x minus the tangent of x equals zero. So my approach here is going to be to factor. Let's factor out one of those tangents.
So I have the tangent of x times the tangent of x minus 1. So remember that when two things are multiplied to equal 0 we have two options either one or both of those factors could be equal to 0. So I have these two equations to solve. So where is the tangent zero? So the tangent is sine over cosine.
So in order for that to be zero, the... and maybe I should say let's just look at the interval zero to two pi. So where is the tangent 0?
The tangent is the sine over the cosine. So in order for the tangent to be 0, the top has to be 0. So the sine has to be 0, and the sine is 0 at 0 and 2 pi. All right, our y coordinate is going to be 0 at these two values.
Now for the second factor, we'll add the 1 over tangent of x equals 1. So here, where is the tangent equal to 1? So let's go back to the unit circle. The tangent is going to be equal to 1 when the sine and the cosine are the same. Right, sine divided by cosine, root 2 over 2 divided by root 2 over 2 is 1. So here's one solution, pi over 4. And then also over here in quadrant 3, 5 pi over 4. The tangent is also going to be 1. Sine divided by cosine is 1 in both of these locations.
So we have two more solutions. We have the solution pi over 4 and 5 pi over 4. So this equation actually has four solutions inside of that interval 0 to 2 pi. And if you will please allow me to back up just a little bit.
There's actually a third Solution here that I just realized I forgot the tangent is going to be 0 at 0 Pi and 2 Pi and so it turns out that we actually have five solutions not four So for the inverse trig functions if I'm thinking about the trig function itself The standard trig function the trig function accepts an angle and returns a ratio So if I wanted a function that took a ratio and returned an angle, well that would be the inverse. So I would need to describe an inverse trig function. However, we sort of have mentioned we've...
I sort of alluded to that there's an issue here. Trig functions are periodic, and periodic functions are not one-to-one. So how do we get over that fact and be able to describe some kind of inverse trig function?
So the answer is that we will force the function to be one-to-one by imposing a domain restriction. So here's the sine function. So very clearly, it's not one-to-one on its domain.
So its domain is all real numbers. You'll notice that any horizontal line is passing through multiple times, which means the outputs are being repeated multiple times. And in fact, the outputs are being repeated infinitely.
So this function is clearly not one to one. However, on this segment that I have, have highlighted in red, this segment spans from negative 1 to positive 1. And our interval here is negative pi over 2 to positive pi over 2, the closed interval. So in this slice of the curve, or this piece, of the curve highlighted in red, we are 1 to 1. And so we can define an inverse by restricting the domain in a way that forces the function to be 1 to 1. So what we have is what we call the inverse sine function. So we write f of x equals the arc sine of x with the domain negative 1 to positive 1 and the range negative pi over 2 to positive pi over 2. So the domain restri- restriction, we're making, we're completing, we're describing an inverse, so the domain becomes the range, and the range becomes the domain. So trig functions are not one-to-one, so they don't have a natural inverse, but I can force them to have an inverse by forcing them to be one-to-one on a particular domain.
So notationally, we often, you'll often see this minus one notation notation for inverses, but these are equivalent. And since it's easy to make this mistake that this is an exponent, my preference is to just stick with the arc notation since there is no possibility of confusing this particular notation as an exponent. It's a notation that's common for inverses, but often confused as an exponent. And so I will typically stick to the arc notation. But you should be able to recognize both of them.
So here's a graph of the inverse sine function. So here is that segment from negative pi over 2 to positive pi over 2 highlighted in red. And so to get the graph of the inverse, we take that process of reflecting about the line y equals x. So in blue here is the graph.
of the inverse sine function. Same thing happens for inverse cosine. So for the inverse cosine, our restriction is zero to pi.
And so we reflect about the line y equals x. Here in blue is the graph of the inverse cosine. And for the tangent function, we make a restriction from negative pi over two to positive pi over two.
That's an open interval, so we have parentheses on the end. We reflect about the line y equals x and right here in blue is going to be the graph of the inverse tangent. The inverse tangent has horizontal asymptotes at these values of positive and minus pi over 2. So again, all of these inverse trig functions are achieved by forcing the function to be one-to-one through a domain restriction. And so since we are restricting the domain to get an inverse, those restricted domains become our ranges.
All right, so here's the domain and range for the inverse sine, inverse cosine, and inverse tangent. And the same process for the inverse cosecant, inverse secant, inverse cotangent. Again, to me, I think you should not worry about all of this here.
This is a lot to memorize. A lot of really messy notation. If you focus on this here, just the inverse sine, inverse cosine, inverse tangent, I think you'll be fine. So it's important to know the restricted domains because the angle you give as an evaluation of an inverse trig function must be in the range of that restricted trig function. So it's very easy to make mistakes here if we don't know the restricted domains and consequently the restricted ranges.
is. We'll get to that in just a second. These are equivalent expressions. So since these are inverses, these expressions are equivalent. y is the arc sine of x is the same as saying the sine of y is equal to x.
That manipulation might be helpful from time to time. So let's finish up with a few things here applying the inverse trig functions. So let's evaluate these expressions. First off, I have the arc cosine of one half. So So remember that a inverse trig function accepts a ratio and tells you the corresponding angle.
So remember that the restricted domain for the cosine, and hence the range of the inverse, is going to be, inverse cosine is going to be 0 to pi. So the range of this function is 0 to pi. So in my mind, I sort of ask myself the question, on the interval 0 to pi, where is the cosine equal to 1 half?
Right? And what is the cosine? what you should find.
This requires sort of thinking backwards to your unit circle. Where in the interval 0 to pi is the cosine 1 half? That's going to occur at pi over 3. All right, next here, again, another arc cosine evaluation.
So we are looking at the same range. So where in the interval 0 to pi is the cosine negative 1 over root 2? So it may be. help to rationalize.
So negative 1 over root 2 is the same, we'll multiply by root 2 over 2, and this becomes negative root 2 over positive 2. So equivalently, I have the arc cosine of negative root 2 over 2. And so where on the interval 0 to pi is the cosine equal to negative root 2 over 2? So cosine being negative puts us in quadrant 2. And that's going to be the angle 3 pi over 4. All right, here I have the expression the cosine of the arc cosine of negative 1. so let's approach it this way so let's just think about this as being the cosine now on the interval 0 to pi where is the cosine equal to negative 1 that happens exactly at pi so this is just equal to the cosine of pi and the cosine of pi is again negative 1. So notice here that we just effectively had the result of the cosine and the arc cosine washing each other out. And that was possible because the domain of the range of the cosine function is equal it contains that point negative 1 however for this last one here it may be tempting to do the same we have a composition with the inverses so it may be tempting to say that this is simply 7 pi over 6 but the problem here is that 7 pi over 6 is not in the range of the arc cosine function.
The range is 0 to pi, and 7 pi over 6 is outside of that. So I have to approach this a little bit differently. So I'm going to write this as the arc cosine.
we will use the unit circle to evaluate the inside. So the cosine of 7 pi over 6 is going to be negative root 3 over 2. And now I ask myself, in the interval 0 to pi, where is the cosine negative root 3 over 2? And referencing the unit circle, we're going to find that this equals 5 pi over 6. So the angle whose cosine is negative root 3 over 2 is 5 pi over 6. So notice that, indeed, it was not the case that these inverses could just wash each other out. Again, 5 pi over 6 is in the range of 0 to pi.
7 pi over 6. was not. Let's finish up here with this example. So I'd like to use a right triangle to simplify this expression. We're going to assume here that the value of x is larger than zero. So how will we approach this?
Well, remember that the inverse trig accepts an angle, excuse me, accepts a ratio and returns an angle. So I'm going to define things this way. So let's let that angle, we'll call it theta. And we'll define this this way, theta is equal to the arc cosine of x over 2. So this describes a triangle.
This describes a triangle whose cosine is x over 2. Let's draw that. So the cosine being the adjacent over the hypotenuse. So we have just one missing side here.
Maybe we'll call it y. So I have y squared plus x squared is equal to 2 squared. Obviously, y squared plus x squared equals 4. And we'll solve for y.
So y squared is 4 minus x squared. And so we're going to take our, because of the way we've written this here, we'll take this y to be positive. So we have y equals the square root of 4 minus x squared. All right, so then I can write it this way. So then the sine of theta.
So now we have this y on the side is just equal to 4 minus x squared under the square root. And so opposite over hypotenuse 4 minus x squared divided by 2. Alright, but what did we say about theta? Theta was the arc cosine of x over 2, so let's just put it back into the form that we were starting with.
So I can write this now, the sine of the arc cosine of x over 2 is equal to this expression, root 4 minus x squared, divided by 2.