In Exercises 1–8, a point on the terminal side of angle θ is g...

Question

In Exercises 1–8, a point on the terminal side of angle θ is given. Find the exact value of each of the six trigonometric functions of θ. (-1, -3)

Accepted Answer

Hello, today we're gonna be determining the values of all six trigonometric functions with the given point. So in order to find the values of our six trigonometric functions, let's go ahead and create a right triangle with our given point, we are given negative five comma negative 11. And since this is a point with two negative values, this is going to be located in quadrant three of an axis. So if we were to draw a diagram, we're going to draw a axis with quadrant three. And we're going to label our point roughly about here, we can use this point to draw a right triangle and we can place alpha our angle in the center of this triangle. Now, we need to label the three sides of the give triangle. In order to form this triangle, we're going to have a travel distance of negative five. So we can assign the long leg a distance of negative five. And in order to reach the point, we need to travel the height of negative 11 units. So we can assign a height for this triangle as negative 11. So negative five is going to be our X value and negative is going to be our Y value. Now, what we need to label is the length for the hypotenuse, but we don't have a given length for the hypotenuse. Instead, we have to solve for it, we can solve by using the Pythagorean theorem. And that's given to us as hypotenuse squared is equal to X squared plus Y squared. We know the value of X is negative five and we know the value of Y is negative 11. So if we plug this into the Pythagorean theorem, we get H squared is equal to negative five squared plus negative 11 squared, negative five squared is going to give us the value of negative 11 squared is going to give us the value of 25 plus 121 is going to give us 146. So we have hypotenuse squared is equal to 146. Finally, we can solve for the hypotenuse by taking the square root of both sides of the equation. And that is going to give us hypotenuse is equal to the square root of 146. So we can give the hypotenuse a length of the square root 1 46. Now we can go ahead and use the values on this right triangle to solve for six trigonometric functions. Now we have an identity for each of the six trigonometric functions sine is going to equal to the Y value over the hypotenuse of a right triangle, cosine is equal to the X value over the hypotenuse of a right triangle. And tangent is equal to the Y value over the X value. And for Cossin and Cotangent, the easiest way to remember them is to note that they are the inverses of sine cosine and tangent respectfully. What this means is that cos is equal to the hypotenuse over the Y values is equal to the hypotenuse over the X value. And cotangent is equal to the X value over the Y value. So now that we have the identities for the six trigonometric functions and we have the right triangle for alpha, we can go ahead and plug in the values to solve for Sine cosine tangent, Cossin and cotangent. Let's go ahead and start with S sign again. Sine is defined as the Y value of our right triangle over the hypotenuse. The Y value in this case is negative 11 and the hypotenuse is the square root of 1 46. So Sine is going to equal to negative 11 over the square root of 146. But keep in mind that we cannot leave a square root in the denominator. So we're going to need to rationalize this value in order to rationalize we're gonna multiply the numerator and denominator by the square root of 1 46. And this is going to give us a sign value of negative 11 times the square root of 1 46/ 46. Now, let's go ahead and find cosine, cosine is defined as our X value over the hypotenuse. And that means cosine is equal to negative five over the square root of 1 46. And just like sine, we're going to need to rationalize this value. So we're gonna multiply the numerator and denominator by the square root of 46. And that is going to give us a cosine value of negative five times the square root of 1 46 all over 146. Now tangent is defined as our Y value over the X value that is going to give us negative 11 over negative five. And that is going to simplify to give us positive 11/5. Now, for our sequent value coin is defined as our hypo news over the Y value that is going to give us the square root of 1 46 over negative 11. For second seeking is defined as our hypotenuse over the X value. And that is going to give us the square root of 1 46 over negative five. And co tangent is defined as the X value over the Y value that is going to give us negative five over negative 11, which is going to simplify, to give us positive 5/11. Now, to simplify the values that we just calculated, we solved that sign is equal to negative 11 times the square root of 1 46/ 46 cosine is equal to negative five times the square root of 46/1 46 tangent is equal to positive 11/5 cos is equal to the square root of 1 46 over negative second is equal to the square root of 1 46 over negative five. And cotangent is equal to positive 5/11. With that being said, these are going to be the six trigonometric values for the given alpha and the given terminal point. And this is going to be the answer to this problem. So I hope this video helps you in understanding how to approach this type of problem. And I'll go ahead and see you all in the next video.

In Exercises 1–8, a point on the terminal side of angle θ is given. Find the exact value of each of the six trigonometric functions of θ. (-1, -3)

Key Concepts

Coordinates and the Terminal Side of an Angle

Radius (r) and Its Calculation

Six Trigonometric Functions and Their Definitions

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