Find the values of the six trigonometric functions for an angl...

Question

Find the values of the six trigonometric functions for an angle in standard position having each given point on its terminal side. Rationalize denominators when applicable. (6√3 , ―6)

Accepted Answer

Welcome back. I am so glad you're here. We're asked to determine the six trigonometric function values of an angle that has its initial side on the positive X axis and a terminal side passing through the given point, rationalize the denominator when necessary, our given point is negative 10 square at 10. And we recall from previous lessons that when we're given a point, we're first given the X value and then the Y value. So we know that our X is equal to negative 10 square at three and our Y is equal to 10. What else do we know? Well, we're trying to determine the six trig trigonometric function values of the angle that passes through that point. What are the trigonometric function values of an angle? We recall from previous lessons that's going to be our sign of theta, our cosign of data, our tangent of the, our cotangent of theta, our, of the, and our koi can of theta. And we recall from those same lessons that those are equal to for the sign of the, that is Y divided by R. The cosine of theta is X divided by R, the tangent of theta is Y divided by X. The cotangent of theta is X divided by Y. The C can of theta is R divided by X and the co can of theta is R divided by Y. And we already know our XX and our Y, we just don't know our R yet. Well, we recall from previous lessons that R is equal to the square root of X squared plus Y squared. We know our X and our Y we can solve for RR. So we've got the square root of, instead of X, we have negative 10 square at three squared. Then plus instead of Y, we've got 10 squared. Now, for that first one, we're taking negative 10 squared of three squared, we have to square both the negative 10 and the square it of three. So negative 10 squared is a positive 100 that's multiplied by square at three squared, which is just three. That's all still under the radical. And it needs to be added to 10 squared, which is 100. Well, 100 multiplied by three is 300. So we have the square root of 300 plus 100 which is the square root of 400 the square root of is 20. So our R value is 20. Now we can solve for those six trigonometric function values of the angle that pass through our given point. All right. So we've got the sign of theta equals Y divided by RY is 10. That's divided by R which is 2010 divided by 20 simplifies to one half. All right, cosine of theta is X divided by Rx in the numerator is negative 10 square at three. That's divided by R which is 20. Well, again, that negative or the 10 and the 20 are going to simplify. So negative 10 divided by 20 is going to be the negative one half. But don't forget that square at three in the numerator. So this simplifies to be negative square at three divided by two. All right, for our tangent of theta Y which is divided by X which is negative 10 square at three. Well, the tens are going to cancel out. So we're gonna have a one in the numerator divided by the square of three. And then we can put that negative up in the numerator. But we want to rationalize this. So we're going to multiply both our numerator and our denominator by the square it of three. So we'll have in the numerator, we'll have a negative square at three because negative one multiplied by square root of three is negative square root three. And in the denominator, the square root of three multiplied by the squared of three is just three. All right, we've got our tangent. Now, our cotangent cotangent is X divided by YX is negative 10 square root three. That's divided by Y which is 10. Again, those tens cancel out. But this one's nice because we don't have to rationalize. So we're left with just negative square at three. All right. C can't theta R in the numerator, R is 20 divided by XX is negative 10 square root three, the 20 divided by the 10. That's just a two in the numerator. But we'll bring that negative up to the numerator. So we have negative two divided by the square of three time to rationalize again, multiplying the numerator and the denominator by the square of three. And so in the numerator, we've got negative two square up three and that's divided by, in the denominator again, square out of three, multiplied by the square out of three is just three. All right. Last one cos theta equals R in the numerator. That's 20 divided by Y which is 10 and 20 divided by 10 is a positive two. So there are our six solutions for our trigonometric functions of the angle that has passing through our given point of negative 10 square at 3 10. Well done. We'll catch you on the next one.

Find the values of the six trigonometric functions for an angle in standard position having each given point on its terminal side. Rationalize denominators when applicable. (6√3 , ―6)

Key Concepts

Coordinates and the Terminal Side of an Angle

Definition of the Six Trigonometric Functions Using Coordinates

Rationalizing Denominators

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