Trigonometric Functions on Right Triangles - Online Tutor, Practice Problems & Exam Prep

Trigonometric functions, including sine, cosine, and tangent, relate angles to side lengths in right triangles. These functions can be remembered using the mnemonic SOHCAHTOA: sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. Additionally, reciprocal identities introduce cosecant, secant, and cotangent, which are the inverses of sine, cosine, and tangent, respectively. Inverse trigonometric functions allow for finding missing angles, utilizing operations like inverse sine to reverse the trigonometric function. Understanding these concepts is essential for solving various mathematical problems involving triangles.

Introduction to Trigonometric Functions

Hey, everyone, and welcome back. So in earlier videos, we've brushed up on this idea of angles and triangles. 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. In this video, there are three main types of trig functions we're going to talk about, which are the sine, cosine, and tangent. 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. An example of a ratio would be something like 3 fifths or 4 fifths. So these types of fractions or ratios are all that these trigonometric functions are. 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 triangle. Now, how can we go about finding the sine of some angle in this triangle that we'll call theta? To find the sine of this, sine is opposite over hypotenuse. So if you wanted to find the sine 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 3, and then divide that by the hypotenuse, which is always the long side of the triangle or 5. So 3 fifths 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 divide it by the hypotenuse. The adjacent side of this triangle is the side next to the angle which in this case is 4, and then this is divided by the hypotenuse or long side of the triangle which is 5. So the cosine of our angle theta is 4 fifths. 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 3, and then the adjacent side is 4. So 3 fourths 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 SOH CAH TOA. This memory tool, SOH CAH TOA, 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. One more thing I want to mention about the tangent is that the tangent of theta is also equal to the sine of theta divided by the cosine of theta. So if we were to take the sine of theta, which we figured out was 3 fifths, and divide that by the cosine of theta, which we figured out was 4 fifths, notice how the fives would just cancel in this fraction, so all we would end up with is 3 over 4, and 3 over 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 SOHCAHTOA 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. 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 SOHCAHTOA. Now looking at example a we have the sine And sine is opposite over hypotenuse according to SOHCAHTOA. 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 over 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 sohcahtoa, which says that tangent is opposite over adjacent. So if we take this ratio, we can go to our angle x, 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 5. So 12 over 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. When dealing with the cosine, cosine is adjacent over hypotenuse based on this memory tool SOHCAHTOA. What you might think we need to do is look at our angle and then find the adjacent side which would be 5 and divide it by the hypotenuse or the long side which is 13. Now if 5 13ths 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 13ths, 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 12, and then the hypotenuse or the long side is still 13. So 12 over 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 SOHCAHTOA memory tool. Hope you found this video helpful. Thanks for watching, and please let me know if you have any questions.

Given the right triangle below, evaluate $\cos\theta$.

• A. $\cos\theta=\frac{8}{17}$
• B. $\cos\theta=\frac{8}{15}$
• C. $\cos\theta=\frac{15}{17}$
• D. $\cos\theta=\frac{15}{8}$

Given the right triangle below, evaluate $\tan\theta$.

• A. $\tan\theta=\frac35$
• B. $\tan\theta=\frac45$
• C. $\tan\theta=\frac43$
• D. $\tan\theta=\frac34$

Fundamental Trigonometric Identities

Hey everyone. So in the last video, we learned about the 3 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 3 new trigonometric functions known as the cosecant, secant, and cotangent. Now just like the 3 trig functions we learned about in the previous video, these functions also have abbreviated versions of their names. We have the cosecant, secant, and cotangent. Now this might sound really scary and kinda 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 3 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 3 trigonometric functions we've 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 3 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 sine. So that means the cosecant is just a flipped version, meaning cosecant would be one over the sine. 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 3, would go on bottom of the fraction. So that's the idea of the cosecant. Now the secant is a reciprocal identity for the cosine. So that means that the secant of theta is 1cosθ. So if the cosine is adjacent over hypotenuse, then the secant is hypotenuse over adjacent. So going to our triangle, we have the hypotenuse or the long side, which is 5, and then this is divided by the side adjacent to our angle theta, which in this case is 4. So that's the secant of theta. Now it turns out that the tangent of theta also has a reciprocal identity, which is the cotangent of theta. So the cotangent of theta would be 1tanθ. And so if the tangent of theta is opposite over adjacent, then the cotangent of theta would be the side adjacent to our angle, which is 4 divided by the opposite side of the triangle, which is 3. Notice how the fractions for the cosecant, secant, and cotangent are literally just the flipped versions of the sine, cosine, and tangent. 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 theta is sine over cosine of theta. Well, this same rule of reciprocal identities applies to the cotangent of theta. So what we would do is flip this identity here, saying that cotangent is cosine over sine. 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 to 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 SOHCAHTOA. SOHCAHTOA tells us how the 3 main trigonometric functions are related to the sides of a triangle. Now let's start by taking a look at example a. For example a, we are asked to find the secant of our angle x. Notice in this situation, we are with reference to angle x. Now when dealing with the secant, we learned that the secant 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 5 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 secant is just the reciprocal of the cosine, we just flip this fraction. So it would become 13 over 5. So that is how you can find the secant 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. Now we'll go into our angle x, and we can see the side opposite to x is 12, and then this is divided by the hypotenuse or the long side which is 13. So if we wanted to find the cosecant of x, which is just the flipped version of sine, we just flip this fraction and get 13 over 12. So that is the cosecant of our angle x. Now for example c, we're 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 were 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'll go into the angle y, and here the side opposite the angle y is 5, then this is divided by the side adjacent to y, which is 12. And the flipped version of 5 over 12 would be 12 over 5, which is equal to the cosine of y. So the cosine of y is 12 over 5, and that is the answer to example c. So this is the main idea behind these 3 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.

If $\tan\theta=\frac{12}{5}$, find the values of the five other trigonometric functions. Rationalize the denominators if necessary.

• A. $\sin\theta=\frac{12}{13},\cos\theta=\frac{5}{13},\cot\theta=\frac{5}{12},\sec\theta=\frac{13}{5},\csc\theta=\frac{13}{12}$
• B. $\sin\theta=\frac{5}{13},\cos\theta=\frac{12}{13},\cot\theta=\frac{5}{12},\sec\theta=\frac{13}{12},\csc\theta=\frac{13}{5}$
• C. $\sin\theta=\frac{12}{13},\cos\theta=\frac{5}{13},\cot\theta=-\frac{5}{12},\sec\theta=-\frac{13}{5},\csc\theta=-\frac{13}{12}$
• D. $\sin\theta=\frac{5}{13},\cos\theta=\frac{12}{13},\cot\theta=-\frac{5}{12},\sec\theta=-\frac{13}{12},\csc\theta=-\frac{13}{5}$

If $\sin\theta=\frac{\sqrt{17}}{17}$, find the values of the five other trigonometric functions. Rationalize the denominators if necessary.

• A. $\cos\theta=\frac{\sqrt{17}}{4},\tan\theta=\frac14,\cot\theta=4,\sec\theta=\sqrt{17},\csc\theta=\frac{4\sqrt{17}}{17}$
• B. $\cos\theta=\frac{\sqrt{17}}{4},\tan\theta=-\frac14,\cot\theta=-4,\sec\theta=\sqrt{17},\csc\theta=\frac{4\sqrt{17}}{17}$
• C. $\cos\theta=\frac{4\sqrt{17}}{17},\tan\theta=-\frac14,\cot\theta=-4,\sec\theta=\frac{\sqrt{17}}{4},\csc\theta=\sqrt{17}$
• D. $\cos\theta=\frac{4\sqrt{17}}{17},\tan\theta=\frac14,\cot\theta=4,\sec\theta=\frac{\sqrt{17}}{4},\csc\theta=\sqrt{17}$

Introduction to Inverse Trig Functions

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. To understand this concept of an inverse, we need to look at something we're a little more familiar with. For example, exponents. Let's say we have this number 3, and let's say I were to take this 3 and raise it to an exponent of 2. This would give us 32, and 32 = 9. 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 9 will get us back to 3. 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. Let's say that we have a 30-degree angle, and we take the sine of this angle. The sine of 30 degrees will give us 1/2. But is there a way to go backward and get 30 degrees from 1/2? And the answer is yes. What you need to do is take the inverse sine, because if you take the inverse sine of 1/2, this will cancel the sine 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 sine of theta, this is the exact same thing as the arcsine of theta. This is something to be familiar with in case you come 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. Here we have this triangle, and we're asked to write a trigonometric function to represent the angle theta. Let's see how we can do this, and we're going to go by the steps. 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. The way that I'm going to do this is using our memory tool, SOHCAHTOA, because SOHCAHTOA tells us how the 3 main trig functions relate to the sides of a right triangle. What I notice 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 hypotenuse. Looking at these options that we have, it looks like the sine includes the opposite and hypotenuse. So I'm going to use the sine of theta, 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, recall that sine is opposite over hypotenuse. So if I go to our angle theta, the opposite side with respect to this angle is 6, and the hypotenuse or the long side is 13. So this would be the equation for the sine of theta that we're looking for, and that's step 2. Now, for step 3, we're asked to take the inverse on both sides of the equation to isolate the angle. To do this step, what I'm going to do is recognize we have the sine of theta. So, I'm going to take the inverse sine on both sides of this equation. θ = sin-1613. This right here is going to be our angle theta represented as this trigonometric function. This right here is the answer to this problem and what we're looking for for the angle theta.

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. 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.

Given the right triangle below, use the sine function to write a trigonometric expression for the missing angle $\theta$.

• A. $\theta=\sin^{-1}\left(\frac{5}{13}\right)$
• B. $\theta=\sin^{-1}\left(\frac{12}{13}\right)$
• C. $\theta=\sin^{-1}\left(\frac{5}{12}\right)$
• D. $\theta=\sin^{-1}\left(\frac{13}{12}\right)$

Given the right triangle below, use the cosine function to write a trigonometric expression for the missing angle $\phi$.

• A. $\phi=\cos^{-1}\left(\frac{12}{13}\right)$
• B. $\phi=\cos^{-1}\left(\frac{13}{5}\right)$
• C. $\phi=\cos^{-1}\left(\frac{13}{12}\right)$
• D. $\phi=\cos^{-1}\left(\frac{5}{13}\right)$

How to Use a Calculator for Trig Functions

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 sine, 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 button, you'll be taken to a certain menu. And on this menu, you will have an option to choose between degrees or radians. 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 sine 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 calculator. 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 parentheses and then hit enter. Once you click enter on your calculator, round it to the nearest tenth, you should get about 0.6. So 0.6 would be the decimal approximation for what you get from the sine 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 2 pi over 15. Now, notice in this example, we're dealing with pi. And because of this means you actually want to be in radian mode for this example. So, go to mode on your calculator, and switch to radians. Once you've done this, you're going to find the tangent button on your calculator, and then you're going to type in \( 2 \pi \) 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 tenth. 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 secant of 50 degrees. Now I again notice 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 secant button on our calculator. But if you recall from the reciprocal identities, secant is the same thing as one over cosine. So this would be the same thing as taking 1 and dividing it by the cosine of 50 degrees. So, what I'm going to do is put in 1 into my calculator, and then divide it by the cosine of 50. You can then close that parenthesis, and then you could hit enter. And when you hit enter on the calculator, you should get an approximate value of 1.55, but rounded to the nearest tenth, 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 arctangent of 3 fourths. 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 arctangent of something? Well, we actually discussed this in previous videos. The arctangent is the same thing as the inverse tangent. And whenever you're dealing with an inverse trig function, what you wanna 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 wanna do is go to your calculator, and you wanna 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 from here, you can type in 3 divided by 4, 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 tenth. 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.

What is a positive value of A in the interval $\left[0\degree,90\degree\right)$ that will make the following statement true? Express the answer in four decimal places.

$\sin A=0.9235$

• A. $22.5568°$
• B. $67.4432°$
• C. $22.4432°$
• D. $33.5438°$

What is the positive value of P in the interval $\left[0\degree,90\degree\right)$ that will make the following statement true? Express the answer in four decimal places.

$\cot P=5.2371$

• A. $55.8102°$
• B. $34.1898°$
• C. $10.8102°$
• D. $79.1898°$

What is the positive value of $D$ in the interval $\left[0,\frac{\pi}{2}\right)$ that will make the following statement true? Express the answer in four decimal places.

$\sec D=3.2842$

• A. $0.3094$ rad
• B. $1.2614$ rad
• C. $0.4760$ rad
• D. $1.0934$ rad

Example 1

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 2 decimal places. Now the example that we have here is actually a continuation 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 trigonometric function that includes the correct sides and angles, which is the sine. 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 now is 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 notice 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 trigonometric function. This will give you the inverse trigonometric function that you're looking for. And since I can see we're looking for inverse sine, you're going to want to press the second button and then sine on your calculator. Now once you have the inverse sine, you're going to type in the ratio 6 over 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 trigonometric function, and we're asked to round our answer to 2 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. I hope you found this video helpful. Thanks for watching.

Here’s what students ask on this topic:

What are the primary trigonometric functions and how are they defined in a right triangle?

The primary trigonometric functions are sine (sin), cosine (cos), and tangent (tan). They relate the angles of a right triangle to the lengths of its sides. These functions can be remembered using the mnemonic SOHCAHTOA:

• Sine (sin) is defined as the ratio of the length of the opposite side to the hypotenuse: $\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$.
• Cosine (cos) is the ratio of the length of the adjacent side to the hypotenuse: $\frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$.
• Tangent (tan) is the ratio of the length of the opposite side to the adjacent side: $\frac{\mathrm{opposite}}{\mathrm{adjacent}}$.

How do you use the mnemonic SOHCAHTOA to remember trigonometric functions?

SOHCAHTOA is a mnemonic that helps remember the definitions of the primary trigonometric functions:

• SOH: Sine (sin) is Opposite over Hypotenuse: $\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$.
• CAH: Cosine (cos) is Adjacent over Hypotenuse: $\frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$.
• TOA: Tangent (tan) is Opposite over Adjacent: $\frac{\mathrm{opposite}}{\mathrm{adjacent}}$.

This mnemonic helps you quickly recall which sides of the triangle to use for each function.

What are the reciprocal trigonometric functions and how are they related to the primary trigonometric functions?

The reciprocal trigonometric functions are cosecant (csc), secant (sec), and cotangent (cot). They are the reciprocals of the primary trigonometric functions:

• Cosecant (csc) is the reciprocal of sine: $\frac{\mathrm{hypotenuse}}{\mathrm{opposite}}$.
• Secant (sec) is the reciprocal of cosine: $\frac{\mathrm{hypotenuse}}{\mathrm{adjacent}}$.
• Cotangent (cot) is the reciprocal of tangent: $\frac{\mathrm{adjacent}}{\mathrm{opposite}}$.

These functions are useful in various trigonometric calculations and identities.

How do you find a missing angle in a right triangle using inverse trigonometric functions?

To find a missing angle in a right triangle, you can use inverse trigonometric functions. If you know the value of a trigonometric function, you can find the angle by taking the inverse of that function:

• Inverse sine (sin^-1 or arcsin): $\theta =\; sin-1(\frac{\mathrm{opposite}}{\mathrm{hypotenuse}})$.
• Inverse cosine (cos^-1 or arccos): $\theta =\; cos-1(\frac{\mathrm{adjacent}}{\mathrm{hypotenuse}})$.
• Inverse tangent (tan^-1 or arctan): $\theta =\; tan-1(\frac{\mathrm{opposite}}{\mathrm{adjacent}})$.

These functions will give you the angle in degrees or radians, depending on your calculator settings.

How do you use a calculator to evaluate trigonometric functions?

To evaluate trigonometric functions using a calculator, follow these steps:

1. Ensure your calculator is in the correct mode (degrees or radians) based on the problem.
2. For sine, cosine, and tangent, use the respective buttons (sin, cos, tan) and input the angle.
3. For reciprocal functions (csc, sec, cot), use the reciprocal identities: csc = 1/sin, sec = 1/cos, cot = 1/tan.
4. For inverse functions (sin^-1, cos^-1, tan^-1), press the second function button (often labeled as 2nd or shift) followed by the trigonometric function button.

For example, to find sin(37°), ensure the calculator is in degree mode, press sin, enter 37, and hit equals.