In Exercises 1–4, a point P(x, y) is shown on the unit circle cor...

Question

In Exercises 1–4, a point P(x, y) is shown on the unit circle corresponding to a real number t. Find the values of the trigonometric functions at t.

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the trigonometric function values for the real number represented by the point P on the unit circle where the point P is the negative of the square root of three divided by three and the negative square root of six divided by three. And the angle T is the angle formed between the x axis and the point P. Now, what do we want to really know? We, we, we really want to know what the trigonometric function values are. And our trigonometric functions are the sign of the anger, the cosine of the angle and the tangent of the angle. We also have the reciprocals of those where the is the reciprocal of sign. The second is the reciprocal of the cosine and the core tangent is the reciprocal of the tangent. So if we can find the values of the sign the cosine and the tangent, we can reciprocate them to find the values of the co the second and the core tangent respectively. Now, what do we already know? Well recall that on the unit circle? OK, on our unit circle or point has the core sign of the angle as the X value and the sign of the angle as the Y value. OK. So already we know the values of the core sign of T and the sign of T. So we can use those to help us figure out the rest. So that tells us that negative root square root of three divided by three is going to be the value of the cosine. So let me write that here and the negative square root of six divided by three is going to be the value of sine. So we have both of those already or are we going to figure out the tangent? Well, again, recall that the tangent of an angle is the ratio of the sign of the anger to the core sign of the angle. We already know the sign and the core sign. So we can use that to help us figure out the value of the tangent. So in this case, let me put it back in red. That's going to be the sign of the angles negative roots square of six divided by three divided by the core sign of the angle. So that's the negative square root of three divided by three. No here let me rewrite it. We have a negative dividing a negative. So we know our result will be positive. So no, that is going to be the square root of six divided by three multiplied by three divided by the square root of three. No, when we do that, OK, then the three cancels the three and I can rewrite this as the square root of six divided by the square root of three. But remember what we know about square roots, the, the laws of indices because let me put it in a little box over here. Recall that if I have two square roots dividing each other, then it's the same thing as writing their caution under the square root. So this is really just the square root of six divided by three, which is equal to the square root of two. So the tangent of T is equal to the square root of two. So I can come back here and I can write that in my, in my, in my spot. So now that we have the values of the sign, the cosine and the tangent, we can go ahead and find the cos which is the reciprocal of the sign. So that is equal to three. I say the negative three divided by the square root of six. No, I want to simplify it. So I, I need to make some more space. So let me come down here. So solving for first, let's start with the cos of T. So as we said, the consequence of T is negative three divided by square of six. So we're going to rationalize that. So let's multiply it by the square of six both cases. So that is going to be the square root of six multiplied by negative three, negative three, multiplied by the square of six all divided by six, which when we go ahead and cancel or factor out that is going to be negative squared of six divided by two. So that will be the of tea. Yeah, it looks like I really need a lot of space right next, I think I'm gonna write it in one straight line. Next. The equant of T is now going to be equal to the reciprocal of negative root three divided by three. So that is going to be three, a negative three divided by the square of three. Let me rationalize multiply it by the spirit of three. So negative three times a square of three, that's negative three multiplied by root three all divided by three. These will cancel or factor out. And that would just give me a negative square of three. So that's my second negative square root of three. And then finally, the core tangent would be the reciprocal of the square of two. I don't think I have to work that out much. That would just be one divided by the square root of two. So these would represent all of our trigonometric function values at the point P. Thanks for watching. I hope this video helped

In Exercises 1–4, a point P(x, y) is shown on the unit circle corresponding to a real number t. Find the values of the trigonometric functions at t.

Key Concepts

Unit Circle

Trigonometric Functions

Angle Measurement

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