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. (5, -5)

Accepted Answer

Hello, today we're going to be determining the values of all six trigonometric functions of alpha with the given terminal point. Now, in order to get the values of the six trigonometric functions, we need to figure out the values of a right triangle. We can use our terminal point to create an image of a right triangle. So if we draw an axis, we need to first figure out the location of our point. Now what we are given is 13 comma negative 13. This is a point in the form of a positive X value and a negative Y value. And that means this point is gonna lie within the fourth quadrant of our axis. We can then use this point to create a right triangle and alpha is gonna be located at the center of this triangle. Now, we need to label the three sides of a right triangle. We know that the distance or the length of this triangle is defined by the X value of our point. So that means the long side or the X value is given to us as positive 13. Now the height of this triangle or the Y value of the coordinate is given to us as negative 13. So we know that the height or Y is equal to negative 13. Now we need to label the length of the hypotenuse of the right triangle. We are not given the hypotenuse, but we do have an equation that can help us solve for it. If we use the Pythagorean theorem, the Pythagorean theorem is defined as hypotony squared is equal to X squared plus Y squared. We know that X is equal to positive 13 and Y is equal to negative 13. So if we plug this into the Hy Pythagorean theorem, we get hypotenuse squared is equal to 13 squared plus negative 13 squared, 13 squared is going to give us the value of 169. A negative 13 squared is going to simplify to give us 169 169 plus 169 is going to give us 338. So H squared is equal to 338. Finally, we're going to take the square root of both sides of this equation and that is going to give us H is equal to the square root of 338. And that is going to simplify to give us 13 times the square root of two. So our hypotenuse is going to have a length of 13 square root of two. Now that we have a right triangle with the three sides labeled, we can go ahead and use these values to determine the values of the six trigonometric functions. So let's first list the identities of the six trigonometric functions for a right triangle. We know that sine is equal to the Y value over the hypotenuse cosine is equal to the X value over the hypotenuse. And tangent is equal to our Y value over the X value is the inverse of sign. And that is given to us as the hypotenuse over why? Second is the inverse of cosine, which is the hypotenuse over X and cotangent is the inverse of tangent which is going to be X over Y. So now that we have the six identities listed, let's go ahead and use the values of the right triangle to figure out the values of sine cosine tangent sequent second and cotangent. So let's go ahead and start with sign. Now again, sine is defined as our Y value over the hypotenuse. This means that that is going to give us the value of negative 13/13 square root two. This is going to simplify to give us negative one over the square root of two. But keep in mind that we cannot leave a square root in the denominator. So we're going to have to rationalize this value by multiplying the numerator and denominator by the square root of two. This is going to simplify and give us assigned value of negative square root of 2/2. Now let's find the cosine value cosine is defined as our X value over the Hypo news. This means that this is going to give us positive 13/13 square root of two. This is going to simplify to give us one over the square root of two. And we're gonna have to rationalize this value as well. We multiply the numerator and denominator by the square root of two. And that is going to leave us with positive square root of 2/2. Now let's figure out the tangent value. Now tangent is defined as our Y value over the X value. So tangent is going to give us negative 13/13 and this is going to simplify to give us negative one. Next, let's figure out coin is defined as the hypotenuse over the Y value. And that is going to give us 13 square root of two over negative 13. And that is going to simplify to give us negative square root of two. Now we figure out our seeking value and seeking is defined as our hypotenuse over the X value. So that is going to give us 13 square root of 2/13, which is going to simplify to give us the square root of two. Now we just need the cotangent value and cotangent is defined as X over Y which is going to be positive 13 over negative 13, which is going to simplify to give us negative one. So let's go ahead and summarize the six values that we just calculated. We know that sine is going to equal to negative square root 2/2 cosine is equal to positive square over 2/2. And tangent is equal to negative one is equal to negative square root of two second is equal to positive square root two and cotangent is equal to negative one. And these are going to be the six values of the trigonometric functions for alpha with the given terminal point. And with that being said, this is going to be the answer to this problem. So I hope this video helps you in understanding how to approach this 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. (5, -5)

Key Concepts

Coordinates and the Terminal Side of an Angle

Definition of the Six Trigonometric Functions

Sign of Trigonometric Functions Based on Quadrants

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