In Exercises 19–24,
a. Use the unit circle shown for Exercises 5...

Question

In Exercises 19–24, a. Use the unit circle shown for Exercises 5–18 to find the value of the trigonometric function. b. Use even and odd properties of trigonometric functions and your answer from part (a) to find the value of the same trigonometric function at the indicated real number. tan 5𝜋/3

Unit circle with coordinates and angles for trigonometric functions in trigonometry course.

Accepted Answer

Hello everybody I have doing. All right. Today, we're going to look at this question that states the circle with a radius of one unit is partitioned into 12 equal arcs where each arc corresponds to specific T values determine the value of the trigonometric function at the given real T real number T using the XO comma Y coordinates in the diagram or specify that the expression is undefined. Also find out the value in part B using even and odd properties of trigonometric functions. Now I'm going to scroll down a bit. So we have a full picture of our graph as well as our answers. And now we can clearly see all of it. Now we see that the graph provided is a unit circle as I mentioned with the radius of one with certain values assigned to it depending on the trigonometric function. Now, in part A, they're asking us to find out the tangent of A seven pi divided by six. And in part B, they're asking us to find out the tangent of negative seven pi divided by six. Now to A as part A and part B as negative square root of two divided by two as choice B has both part A and B as negative one half as a choice C has both has part A as positive square root of thir of three divided by three and part B as negative square root of three divided by three. And answer choice D has part A as positive squared of two divided by two and part B as negative squared of two divided by two. Now, when looking at this question, we'll go part by part first talking about part A and something important that we have to know is that when it comes to trigonometric uh equations associated with the unit circle, specifically, we're gonna have X is equal to cosine of theta. So when we talk about cosine, we're talking about X values and Y is equal to sine of beta. So when we talk about S sign, it's gonna be associated with the Y. Now what we're dealing with in part A, they told us that we are at seven pi divided by six. And when we look at the unit circle that you provided us, we see that there is a seven pi divided by six and there is an associated point with it. And that point is negative squared of three divided by two comma negative one half. So using that point is how we're going to figure out the tangent. So because we have that point, we will therefore have that the tangent of seven pi divided by six is equal to. Now remember that tangent is sine divided by cosine. So we'll have that tangent of seven pi divided by six is equal to sign of seven pi divided by six, divided by cosine of seven pi divided by six. And we can just plug in the values that they gave us. So for sine of seven pi divided by six and the numerator we'll have negative one half and that will be divided by cosine of seven pi divided by six, which we just look at the X value as we said earlier. And that X value here is negative squared of three divided by two. Now, in this case, the twos will cancel out and will be left with negative one divided by the square roots of three or well negative one divided by the negative squared of three, which means that tangents of seven pi divided by six is equal to because we don't want, well, first of all, the negatives will cancel out and we also don't want a square root in the denominator. So we're going to multiply both the numerator and the denominator by the squared of three to get that the tangent of seven pi over divided by six is equal to the square root of three divided by three. And that will be our answer for part A and then we can move on to part B where we have tangent of negative seven pi divided by six. Now, we should note that tangent of negative theta is equal to negative tangent of data. So using that knowledge, we will therefore have negative tangents of seven pi divided by six. And that will be equal to, we already have tangents of 75 divided by six. So we just throw the negative in front of it and we'll have that, this will be equal to negative square root of three divided by three. And that will be our answer for part B. And if we look at our answer choices, this matches with answer choice C. So as you can see, just by simply using the unit circle and relationships with the trigonometric functions, we are able to solve both parts. So I hope that was helpful. And until next time.

Key Concepts

Unit Circle

Trigonometric Functions

Even and Odd Properties

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