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Math 104: Calculus14 chapters | 116 lessons | 11 flashcard sets

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Lesson Transcript

Instructor:
*Eric Garneau*

Explore how the Pythagorean Theorem can be used in conjunction with trigonometric functions. In this lesson, take an inverse trigonometric function, and define all three sides of a right triangle.

Let's take one more look at trigonometry. Remember there are two things that you need to keep in mind with respect to trigonometry. You need to remember **SohCahToa** - sin(*theta*) equals the opposite over the hypotenuse, cos(*theta*) equals the adjacent over the hypotenuse, and tan(*theta*) equals the opposite over the adjacent. So here's our right triangle. We've got *theta*, the adjacent leg, the opposite leg and the hypotenuse.

The second thing you need to remember is the **Pythagorean Theorem**. That says that if you have this *abc* right triangle, *a*^2 + *b*^2 = *c*^2. With those two things, you can do a lot of important calculations in calculus and geometry.

What are some of the things you might need to know? Well, 1/sin(*theta*) is known as csc(*theta*). It's equal to the hypotenuse over the opposite side. We will never write this as sin^-1(*theta*). Why is that? Well, sin^-1(*theta*) is really the inverse function of sin(*theta*); it's not 1/sin(*theta*). So what this means is that sin^-1(sin(*theta*)) will give you *theta*, just like *f*^-1(*f(x)*) will give back *x*. That's the definition of the inverse function here. This also means that if sin(*theta*) is the opposite over the hypotenuse, the cosecant sin^-1 of the opposite over the hypotenuse is equal to *theta*.

Alright, so what's an example of using the Pythagorean Theorem, sines and cosines in a meaningful way? Let's say you have the function sin(*y*) = *x*. This might come from having the equation csc(*x*) = *y*. That's like saying sin^-1(*x*) = *y*. So sin(*y*) = *x*. Sin(*y*) also equals the opposite divided by the hypotenuse. So let's draw out a right triangle and let's make our angle *y*. Here I've got the opposite side and here I've got the hypotenuse. I know that the opposite divided by the hypotenuse is equal to *x*, so why don't I just call the hypotenuse 1 and this opposite side equal to *x*? Opposite over hypotenuse is equal to *x* divided by 1, so this triangle makes sense with our equation sin(*y*) = *x*.

From the Pythagorean Theorem we can find out what this other side equals. I know that *x*^2 plus the length of this side squared has to equal 1. That means that this side is equal to the square root of 1 - *x*^2, so let's put that in our triangle. Now this triangle represents sin(*y*) = *x*. Now that we know that, we can find out what the cosine and tangent of *y* are. The cosine of *y* is equal to the adjacent side divided by the hypotenuse. The adjacent side is the square root of 1 - *x*^2, and the hypotenuse is just 1, so here's my cosine of *y*. Similarly, the tangent of *y* equals the opposite over the adjacent, which is just *x* divided by the square root of 1 - *x*^2. All of these represent this right triangle.

You can do the same thing if you have cos(*y*) = x. This might have come from sec(*x*) = *y*, so if cos(*y*) = *x* that's like saying *x* is equal to the adjacent over the hypotenuse. So let's make the hypotenuse 1 again, and the adjacent side equal to *x*. The opposite side, by the Pythagorean Theorem, equals the square root of 1 - *x*^2, and I can find sine of *y* and tangent of *y* based solely on this.

This is going to become very useful if you can remember it throughout all of calculus. It's really just two rules: 1) SohCahToa and 2) the Pythagorean Theorem. Remember these and you're set.

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Math 104: Calculus14 chapters | 116 lessons | 11 flashcard sets

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