Trigonometric Functions on Right Triangles - Video Tutorials & Practice Problems
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1
concept
Introduction to Trigonometric Functions
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6m
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Hey everyone and welcome back. So in earlier videos, we've brushed up on this idea of angles and triangles and what we're going to be looking at in this video is arguably one of the most critical topics both in this course and in future math courses that you will likely take, which is the idea of trigonometric functions, which we also call trig functions for short. Now, if you're not sure what trig functions are, all these do is they relate angles to the side lengths of a right triangle. Now, in this video, there are three main types of trig functions we're going to talk about, which is the sine cosine and tangent. Now that might sound a little confusing or like it's a bunch of gibberish but don't sweat it because all we're going to learn is that these three trigonometric functions are just ratios. And an example of a ratio would be something like 3/5 or 4/5. So these types of fractions or ratios are all that these trigonometric functions are. So without further ado let's get into this video and see how we can solve some problems where we need to find these specific ratios based on a right angle. Now, how could we go about finding the sign of some angle in this triangle that we'll call theta? How can we find the sign of this wells is opposite over hypotenuse? So if you wanted to find the sign of this angle, theta what you do is look at the angle that you have and go to the opposite side of the triangle, which in this case is three and then divide that by the hypotenuse, which is always the long side of the triangle or five. So 3/5 is going to be the sine of theta. As you can see, solving for these trigonometric functions is pretty straightforward. So let's now say we wanted to find the cosine of theta. Well, if we're dealing with the cosine this time, what we need to do is look at our angle theta and find the adjacent side and divided by the hypotenuse. Now the adjacent side of this triangle is the side next to the angle which in this case is four. And then this is divided by the hypotenuse or long side of the triangle, which is five. So the cosine of our angle theta is 4/5. But now let's say we wanted to find the tangent of our angle theta. If we're dealing with the tangent this time, what we need to do is take the opposite side of the triangle and divide it by the adjacent side. So the opposite side to our angle, theta is three and then the adjacent side is four. So 3/4 would be the tangent of our angle theta. Now, you might feel like this is a lot to remember like how are we just supposed to know that sine is opposite over hypotenuse or that tangent is opposite over adjacent? Well, it turns out there is a memory tool that we use to remember this and that is so Koa because this memory tool SOCA Towa tells us how each of the trigonometric functions relates to the sides of the right triangle. So we have that sine is opposite over hypotenuse cosine is adjacent over hypotenuse and tangent is opposite over adjacent. Now, one more thing I want to mention about the tangent is that the tangent of theta is also equal to the sin of the divided by the cosine of theta. So if we were to take the sin of the, which we figured out was 3/5 and divide that by the cosine of data, which we figured out was 4/5 notice how the fives would just cancel in this fraction. So all we would end up with is 3/4 and 3/4 is exactly what we calculated the tangent to be. So this is another relationship to be familiar with when dealing with these trigonometric functions. Now to really make sure we have these trig functions. And this Soko Towa memory tool down let's try some examples where we have to solve some problems based on a right triangle. Now, in this problem, we are asked to find the values of the trig function indicated given the right triangle that we have over here. Now, we're going to start with example A and before I do anything, what I'm going to do is write down the memory tool Soko Towa. Now looking at example A, we have the sign and S is opposite over hypotenuse according to Sokoto. So we'll have the opposite side of our right triangle divided by the hypotenuse. Now we're dealing with angle X in this example. So if we go to angle X in our triangle, we're going to have the side opposite to angle X which is 12 divided by the hypotenuse or the long side which is 13. So 12/13 is the ratio we're looking for. And the answer for example, A now let's try solving example B where we are asked to find the tangent of X. Well, if we once again go to our angle X here, we can find the tangent by looking at Soko Towa which says that tangent is opposite over adjacent. So if we take this ratio, we can go to our angle X and we can find the side opposite to X which in this case is 12 and then divide it by the adjacent side or the side next to X, which in this case is five. So 12/5 is the ratio that we're looking for. For example, B. But now let's try solving example C where we are asked to find the cosine of our angle Y. Now, when dealing with the cosine cosine is adjacent over hypotenuse based on this memory tool. So Koa, now what you might think we need to do is look at our angle and then find the adjacent side which would be five and divided by the hypotenuse or the long side, which is 13. Now, if 5/13 is what you're thinking, the correct ratio is this is actually not correct. And here's the reason why notice in this example, we are dealing with angle Y and because we're dealing with angle Y, we can no longer use X as a reference. We now have to use Y as the reference angle. So rather than having 5/13 we're going to have the ratios based on the adjacent and hypotenuse of Y. So if we go to Y, the adjacent side to Y is actually 12 and then the hypotenuse or the alongside is still 13. So 12/13 would be the ratio for the cosine of Y. So this is something you need to watch out for when dealing with these types of problems. Because notice in the first two examples, we were dealing with angle X. So we used X as a reference where in this example, we were dealing with Y. So we used angle Y as a reference. So that's the main idea behind solving trigonometric functions and using this Sokoto a memory tool. Hope you found this video helpful. Thanks for watching. And please let me know if you have any questions.
2
Problem
Problem
Given the right triangle below, evaluate cosθ.
A
cosθ=178
B
cosθ=158
C
cosθ=1715
D
cosθ=815
3
Problem
Problem
Given the right triangle below, evaluate tanθ.
A
tanθ=53
B
tanθ=54
C
tanθ=34
D
tanθ=43
4
concept
Fundamental Trigonometric Identities
Video duration:
5m
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Video transcript
Hey, everyone. So in the last video, we learned about the three main types of trigonometric functions, which are the sine cosine and tangent. And what we're going to be learning about in this video is three new trigonometric functions known as the co sequent sequent and cotangent. Now, just like the three trig functions we learned about in the previous video, these functions also have abbreviated versions of their names. We have the cosecant sequent and cotangent. Now this might sound really scary and kind of complicated because there's just like a lot to remember and a lot of different things that we've learned about so far, but don't sweat it with these three trigonometric functions because what we're going to learn in this video is that these three functions are actually very closely related to the last three trigonometric functions we learned about in a very simple way. So without further ado let's get right into things to see what these new trig functions are all about. Now, the way that these functions are related to the previous three trig functions is they are reciprocals of the other trig functions. Now, to be specific, we call these reciprocal identities, but all a reciprocal identity is is it's just a fancy way of saying that this thing is a flipped version of that thing. Now, to understand this a little bit better, let's take a look at the cosecant. For example, the cosecant is a reciprocal identity for the. So that means the cosecant is just a flipped version, meaning cosecant would be one over the sign. So if the sine of theta is opposite over hypotenuse, the cosecant of theta would be hypotenuse over opposite. So going to a right triangle, for example, we can look at our angle theta and the hypotenuse or the long side would go on top of the fraction and then the side opposite to the angle which in this case is three would go on bottom of the fraction. So that's the idea of the cosecant. Now the sin is a reciprocal identity for the cosine. So that means that the sin of data is one over the cosine of data. So if cosine is adjacent over hypotenuse, then see it is hypotenuse over adjacent. So going to our triangle, we have the hypotenuse or the long side which is five and then this is divided by the site adjacent to our angle theta which in this case is four. So that's the secret of theta. Now it turns out that the tangent of data also has a reciprocal identity, which is the cotangent of data. So code tangent of data would be one over the tangent of data. And so if the tangent of data is opposite over adjacent, then the cotangent of data would be the side adjacent to our angle, which is four divided by the opposite side of the triangle, which is three. Notice how the fractions for the coss and cotangent are literally just the flipped versions of the sine cosine and tang. So this is the main idea and concept behind reciprocal identities. Now something else I do want to mention is we talked about this identity in the previous video, how the tangent of data is sign over cosine of data. Well, this same rule of reciprocal identities applies to the cotangent of data. So what we would do is flip this identity here saying that cotangent is cosine over S sign. So this is just another identity to watch out for when dealing with these types of trigonometric functions. Now to really make sure we have this concept solidified, let's actually try some examples where we have to use these new reciprocal identities. So in these problems, we're told find the value of the trig function indicated given this right triangle. Now, before I do anything, what I'm going to do is write down the memory tool we learned about in the previous video, which is Sokoto Sokoto tells us how the three main trigonometric functions are related to the size of a triangle. Now, let's start by taking a look at example, a for example, a we are asked to find the sequence of our angle X notice in this situation, we are with reference to angle X. Now, when dealing with the sequent, we learn the sequin is related to the cosine and the cosine of X is going to be equal to adjacent, divided by hypotenuse. Now, if we go to our triangle and look at our angle X, the side adjacent to X is five and then this is divided by the hypotenuse or the longest side, which is 13. So this is what the cosine is equal to. And since the second is just the reciprocal of the cosine, we just flip this fraction. So it would become 13/5. So that is how you can find the sequence of angle X using these reciprocal identities. But now let's take a look at example, B where we are asked to find the cosecant of X. Well, we know that the cosecant is related to the sine. So we have the sine of X and the sine of X is going to be equal to opposite over hypotenuse. We'll go into our angle X, we can see the side opposite to X is 12. And then this is divided by the hypotenuse or alongside which is 13. So if we wanted to find the cosecant of X, which is just the flipped version of sign, we just flip this fraction and get 13/12. So that is the cosecant of our angle X. Now, for example, C, we are asked to find the cotangent of Y but notice this time we're looking at angle Y so we can no longer be with reference to X. So this is something we talked about in the previous video, you wanna make sure you are paying attention to which angle you are with reference to. So since we're with reference to angle Y, we can find the cotangent of Y by recognizing that the cotangent is the reciprocal for the tangent. So we have the tangent of Y we can see here is equal to opposite over adjacent, we're going to angle Y here, the side opposite to angle Y is five, then this is divided by the side adjacent to Y which is 12 and the flipped version of 5/12 would be 12/5, which is equal to the cosine of Y. So the cosine of Y is 12/5. And that is the answer to example, C so this is the main idea behind these three new trigonometric functions and how you can solve problems with them using these reciprocal identities. So hope you found this video helpful. Thanks for watching and please let me know if you have any questions.
5
Problem
Problem
If tanθ=512, find the values of the five other trigonometric functions. Rationalize the denominators if necessary.
Hey, everyone and welcome back. So up to this point, we've been spending a lot of time talking about trigonometric functions and how they relate to the sides of a right triangle. But something we haven't really covered yet is how we can deal with finding a missing angle for a right triangle. And that's what we're going to talk about in this video because you can actually find these missing angles using something known as inverse trigonometric functions. So without further ado let's learn about these operations that allow us to find these missing angles. Now, for some problems that you're going to be given, you will have the value of a trigonometric function. And when this happens, it's actually pretty easy to find the missing angle because all you need to do this is the inverse of whatever trig function you have. Now to understand this concept of an inverse, we need to look at something we're a little bit more familiar with. So let's say exponents, for example. So here we have this number three and let's say I were to take this three and raise it to an exponent of two, this would give us three squared and three squared is equal to nine. But is there a way I can undo or reverse this operation that we just did? And the answer is yes, I can, I can take the square root because the square root undoes the square, meaning that the square root of nine will get back to three. So that's the idea of an inverse, the inverse undoes the operation you started with. Now, when it comes to trigonometry, we also have inverse operations known as inverse trig functions. So let's say that we have a 30 degree angle and we take the sign of this angle, the sign of 30 degrees will give us one half. But is there a way to go backwards and get 30 degrees from one half? And the answer is yes, what you need to do is take the inverse sign because if you take the inverse sign of one half, this will cancel the sign operation that you took, meaning you're going to end up with your angle, which is 30 degrees. So that's the idea of an inverse trig function. It undoes the trig operation that you started with. Now something else I want to mention is these inverse trigonometric functions are also called arc functions. So if you see the inverse sign of data, this is the exact same thing as the arc sign of data. This is something to be familiar with. In case you across these arc functions, these are literally just the same thing as inverse functions. Now to make sure that we understand these concepts, let's actually try an example where we need to find the missing angle of a triangle. So here we have this triangle and we're asked to write a trigonometric function to represent the angle theta. So let's see how we can do this. And we're going to go by the steps. Now, our first step when solving this problem is going to be to choose a trigonometric function which includes the correct angle and sides for this triangle. Now, the way that I'm going to do this is using our memory tool, Soko Towa, because Soko Towa tells us how the three main trig functions relate to the sides of a right triangle. Now, what I noticed is that we have our angle theta and with respect to this angle, we have the opposite side of the triangle as well as the long side or the hypotenuse. Looking at these options that we have, it looks like the sign includes the opposite and hypotenuse. So I'm going to use the sign of the and that is going to be our first step. Now, our next step is going to be to write an equation with the chosen trigonometric function and to write an equation. Well, recall that s is opposite over hypotenuse. So if I go to our angle, theta, the opposite side with respect to this angle is six and the hypotenuse or the long side is 13. So this would be the equation for the sin of theta that we're looking for. And that's step two. Now, for step three, we are asked to take the inverse on both sides of the equation to isolate the angle. So to do this step, what I'm going to do is recognize we have the sign of the. So I'm going to take the inverse sign on both sides of this equation. The inverse sign is going to allow us to cancel the S operation leaving us with just theta and this is going to be equal to the inverse sine of the fraction that we have and the fraction that we have is 6/13. So this right here is going to be our angle theta represented as this trigonometric function. So this right here is the answer to this problem and what we're looking for for the angle theta. And in future videos, we're actually going to learn how you can take these inverse trigonometric functions and how you can plug them into a calculator to get an approximate value for this angle. But for now this is all we're asked to find for this example. So that is the solution. So hopefully you found this video helpful. This was a basic introduction to inverse trigonometric functions and finding missing angles of a right triangle. Let me know if you have any questions. Thanks for watching.
8
Problem
Problem
Given the right triangle below, use the sine function to write a trigonometric expression for the missing angle θ.
A
θ=sin−1(135)
B
θ=sin−1(1312)
C
θ=sin−1(125)
D
θ=sin−1(1213)
9
Problem
Problem
Given the right triangle below, use the cosine function to write a trigonometric expression for the missing angle ϕ.
A
ϕ=cos−1(1312)
B
ϕ=cos−1(513)
C
ϕ=cos−1(1213)
D
ϕ=cos−1(135)
10
concept
How to Use a Calculator for Trig Functions
Video duration:
4m
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Video transcript
Hey, everyone. So up to this point, we've been spending a lot of time talking about trigonometric functions as well as inverse trigonometric functions. And in this video, we're going to be tying these concepts together by talking about how we can use a calculator to solve problems that involve these types of functions. Now, something that I will say is that for certain problems, you're actually going to have to use your calculator to evaluate the function rather than just using the fraction or ratios that we learned about before. Now, this can be pretty complicated, especially when you're not too sure how to use your calculator properly. But that's what we're going to be going over in this video. And I'm going to be walking you through the steps that you need to take to evaluate these types of trigonometric functions and solve these problems. So without further ado let's get right into things. Now, for trig functions, you're going to need to use the sign cosine and tangent buttons on your calculator, which should look something like this. Now, when dealing with problems where you have to evaluate the trig function on a calculator, you need to make sure you're in the correct mode. Now, the mode button is something that should also be on your calculator and when you press the mode, but you'll be taken to a certain menu and on this menu, you will have an option to choose between degrees or radiance. Now, depending on which of these you choose, is dependent on the type of problem that you're solving. But you need to know when to be in the correct mode for the type of problem. So let's actually take a look at some examples to make sure that we know how to do this. So here we have an example which says, find the value of each of the following trigonometric operations and round to the nearest 10th and keep in mind that the nearest 10th would be the nearest number or the first number after the decimal place. Now let's start with example a where we're asked to find the sign of 37 degrees. Now notice in this example, we have a degree sign and because we have degrees, that means that we want to be in degree mode on our calculus. So once you've switched to degree mode, you're going to type in the sign, which is a button on your calculator. And then what you're going to do is put 37 within these parentheses, you're going to close the parenthesis and then hit enter, once you click, enter on your calculator rounded to the nearest 10th you should get about 0.6. So 0.6 would be the decimal approximation for what you get from the sign of 37 degrees. And that's example A but now let's take a look at example, B where we're asked to find the tangent of two pi over 15. Now notice in this example, we're dealing with pi and because of this, that means you actually want to be in radiant mode for this example. So go to mode on your calculator and switch to radiance. Once you've done this, you're going to find the tangent button on your calculator. And then you're going to type in two pi and then divide that by 15, you're then going to close the parenthesis and press enter. Once you press enter, you should get about 0.4 rounded to the nearest 10th. So that's going to be the answer. For example, B now let's take a look at example, C where we're asked to find the sequent of 50 degrees. Now, I again noticed that we have a degree symbol here which means we want to be in degree mode for this example. Now, the problem with this situation is we don't really have a secret button on our calculator. But if you recall from the reciprocal identities secret is the same thing as one over cosine. So this would be the same thing as taking one and dividing it by the cosine of 50 degrees. So what I'm going to do is put in one into my calculator and then divide it by the cosine of 50. You can then close that parenthesis and then you can hit, enter and when you hit, enter on the calculator, you should get an approximate value of 1.55. But run to the nearest 10th, we can say that that's 1.6. So 1.6 would be the answer. For example, c Now let's try example, d where we're asked to find the arc tangent of 3/4. Now we are asked to find our answer in degrees. So we want to be in degree mode on our calculator. But the question becomes, how can we find the arc tangent of something? Well, we actually discussed this in previous videos, the arc tangent is the same thing as the inverse tangent. And whenever you're dealing with an inverse trig function, what you want to do is you want to press the second button on your calculator. This is a button that you should see. Once you press the second button, you're then going to press the associated trigonometric function. And this should give you the inverse of that trig function. So what you want to do is go to your calculator and you want to press the second button and then once you've pressed second, you're going to press on tangent, this will give you the inverse tangent. And here you can type in three divided by four and then close the parenthesis. Once you've done this, you can hit, enter on this and you should get an approximate value of 36.9 degrees rounded to the nearest 10th. So 36.9 degrees is the answer. For example, D so this is how you can evaluate trigonometric and inverse trigonometric functions on your calculator. So hope you found this video helpful. Thanks for watching and let me know if you have any questions.
11
Problem
Problem
What is a positive value of A in the interval [0°,90°) that will make the following statement true? Express the answer in four decimal places.
sinA=0.9235
A
22.5568°
B
67.4432°
C
22.4432°
D
33.5438°
12
Problem
Problem
What is the positive value of P in the interval [0°,90°) that will make the following statement true? Express the answer in four decimal places.
cotP=5.2371
A
55.8102°
B
34.1898°
C
10.8102°
D
79.1898°
13
Problem
Problem
What is the positive value of D in the interval [0,2π) that will make the following statement true? Express the answer in four decimal places.
secD=3.2842
A
rad
B
rad
C
rad
D
rad
14
example
Example 1
Video duration:
1m
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Video transcript
Welcome back everyone. Let's see if we can solve this example. In this example, we are asked to determine the missing angle theta in degrees for the right triangle below and to approximate our answer to two decimal places. Now, the example that we have here is actually a continuation off of an example that we did in a previous video where we learned about inverse trigonometric functions. Now notice that we have these first three steps complete because we've already chosen a trig function that includes the correct sides and angles, which is the sign we've gone ahead and written the equation and we've taken the inverse on both sides of the equation to solve for our angle theta. So what we need to do is now approximate what this angle theta is based on what we've solved so far. So let's go ahead and see how we can do this. Now, what I noticed is in this example, we are asked to give our answer in degrees. So because of this, you want to make sure your calculator is in degree mode when solving this problem. Now, once you've put your calculator in degree mode, the next step is going to be to press the second key on your calculator and the associated trig function, this will give you the inverse trig function that you're looking for. And since I can see we're looking for inverse sign, you're gonna want to press the second button and then sign on your calculator. Now, once you have the inverse sign, you're going to type in the ratio 6/13 and then close that parenthesis from here, you're going to click enter and this will give you the approximate value for the inverse trig function. And we're asked to round our answer to two decimal places. Now looking at the calculator, your angle theta should be approximately equal to 27.49 degrees. So this right here is the approximate answer for our angle theta and the solution to this problem. So hope you found this video helpful. Thanks for watching.