In Exercises 37–38, a point on the terminal side of angle θ is gi...

Question

In Exercises 37–38, a point on the terminal side of angle θ is given. Find the exact value of each of the six trigonometric functions of θ, or state that the function is undefined.(0, -1)

Accepted Answer

Hello everybody every night today. Today, we're going to be looking at this question that states the given point lies on the terminal side of an angle. Alpha and pro ask they're asking us to provide the uh exact values of sign of alpha cosine of alpha tangents of alpha cotangent alpha C cans alpha and co cans alpha where the point given is zero comma negative 23. Now, when it comes to the trigonometric functions, we are usually dealing with a unit circle. So we have a Y axis and an X axis and then we usually have a circle around that coordinate system and usually the unit circle has a radius of one. Now, in our case, we don't know what the radius is. Now, we know the expression for a circle is X squared plus Y squared equals R squared. Now, if we want to solve for the radius, we'll have that R is equal to the square roots of X squared plus Y squared. So all we have to do now is plug in our values. So we have R is equal to the square roots of from the point that they gave us is zero. So we have the squared of zero squared plus Y squared and they told us why was negative 23. So we'll have negative 23 squared. So we're left with the zero squared will cancel out. And then we'll have negative 23 squared, but we'll have the, the square root of negative 23 squared. So we have the R is equal to positive 23 because once you square the negative 23 the negative will cancel out. And then you take the square root of that positive number which will take us back to 23. So now that we have the radius, we can go ahead and solve for all of our trigonometric functions. Now let's start with sign. So we have that sign of alpha is equal to Y divided by the radius. Now this is equal to negative divided by 23 which is equal to negative one. And that will be our answer for sign of alpha. And we can move on to cosine alpha cosine alpha will be X divided by the radius and this will be equal to zero divided by 23 which is equal to zero. And that will be our answer for the cosine. Then we move on to tangent alpha tangent alpha is equal to Y divided by X and this will be equal to negative 23 divided by zero. And since we can't divide it by zero, we'll have this equal to undefined. And that will be the answer for our tangents. And we can move on to code tangents alpha. Now code tangents of alpha will be equal to one divided by tangents alpha. So we just have the reciprocal of our tangent fraction which is negative zero divided by 23 which is equal to zero because zero divided by anything is zero, then we can move on to cans of alpha C cans of alpha is one divided by cosine of alpha. So we just once again, take the reciprocal of our cosine equation and we'll have 23 divided by zero. And as we said, we can't divide by zero. So we'll have this equal to undefined. And finally, we can move on to cos cans of alpha. Now co sequent of alpha is one divided by sine of alpha. So once again, we take the reciprocal at this time of our sign. But for that, we'll still have negative 23 divided by 23 which is equal to negative one still. So sine and cos can will be the same answer. And just like that by using our knowledge of trigonometric functions with the circle in the coordinate system, as well as the relationship between the functions with that circle, we have been able to calculate their different values. So I hope that was helpful. And until next time

In Exercises 37–38, a point on the terminal side of angle θ is given. Find the exact value of each of the six trigonometric functions of θ, or state that the function is undefined.(0, -1)

Key Concepts

Trigonometric Functions

Unit Circle

Undefined Functions

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