In Exercises 8–13, find the exact value of each expression. Do no...

Question

In Exercises 8–13, find the exact value of each expression. Do not use a calculator.cot (-8𝜋/3)

Accepted Answer

Hello everybody. I have we doing today today. We're going to begin this question that states determine the exact value of the following expression without the help of the calculator where our expression is cotangent of negative seven pi divided by three. Now our answer choices are a squared of three divided by three B zero C negative squared of three divided by three or D undefined. Now, I'm going to rewrite our expression. So we had cool tangents of negative seven pi divided by three. Now, usually you want to have your expression fall within the range of zero and two pie. So you want your data to be between. So zero is less than your theta is less than two pi. Now, in this case, I'm going to rewrite it as the cotangent of negative two pi minus pi divided by three. So I haven't rewritten it as falling within that range. But you'll see why in a second, what I did hear was basically imagine we're going, we have a unit circle or imagine you have a coordinate system with a circle within that coord system. That's what we use for a trigonometric functions. The unit circle Now if you move along that circle, every revolution, all the values are going to repeat. If I move negative seven pi divided by three, that's the same thing as completing negative two pi circles. So you go around the circle one time, 11 whole revolution is negative two pi which would be negative six pi divided by three. So if it gets a negative seven pi divided by three, you subtract another pi divided by three. So we can say that the co tangent of negative seven pi divided by three is equal to the code tangent of negative pi divided by three. And once again, I still have this negative. I haven't said it uh in the range that we wanted because if we recall co tangents of negative A theta is equal to negative cotangents of positive data. So therefore, we can rewrite what we have. So we had code tangents of negative pi divided by three because that's the last part of the final revolution where the the values don't repeat. If you, if you remember what we said, since we're going around in a circle that first revolution we don't really care about because the values repeat. So we care about that last revolution where we don't complete the circle once again. So if we had code tangent of negative pi divided by three, we can rewrite that as negative cotangent of positive pi divided by three. And now the theta falls within the range that I want. So we can move on and say that using the unit circle and these are values that we should have memorized or know how to obtain. So using the unit circle, the unit circle would tell us that pi divided by three has a point of one half comma squared of three divided by two. And we should also know that with the unit circle, the X component is equal to cosine of theta and the Y component of a of a point is equal to sin of theta. So therefore, if we said that we're working with negative code tangents of pi divided by three, that will be equal to negative one divided by tangents of pi divided by three because co tangent of theta is equal to one divided by tangent of theta. Now this will be equal to negative one divided by. Now, tangents of theta is equal to sine theta divided by cosine theta. So we'll have negative one divided by sine of pi divided by three divided by cosine of pi divided by three. And that will be equal to negative one divided by the square root of three divided by two, divided by one half. And we obtained those values from the point that we pointed out earlier since we said that sine of data is the Y for that sign of pi divided by three, we just look at the Y value which was squared of three divided by two. And for the cosine data, we look at the X value which is one half. Now, since we have one divided by another fraction, we just take the reciprocal of that. And we'll have that, this will be equal to negative one half divided by the square root of three divided by two. And in this case, the twos will cancel out and will be left with negative one divided by the square root of three. So therefore, col tangent of negative seven pi divided by three is equal to negative co tangent of pi divided by three, which is equal to, we had negative one divided by the square of three. But we don't want a square root in the denominator. So we multiply both the numerator and denominator by the square of three and we'll be left with negative square root of three divided by three. And that will be our final answer which matches with answer choice C. So as you can see the unit circle is very important when it comes to trigonometric functions. And by knowing the values associated with it, as long as its relationship to the different functions, we're able to solve a question like this. So I hope that was helpful. And until next time.

In Exercises 8–13, find the exact value of each expression. Do not use a calculator.cot (-8𝜋/3)

Key Concepts

Cotangent Function

Angle Reduction

Unit Circle

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