In Exercises 8–13, find the exact value of each expression. Do...

Question

In Exercises 8–13, find the exact value of each expression. Do not use a calculator. sec 22𝜋 3

Accepted Answer

Hello everybody. I hope you're doing. All right. Today, today we're going to be looking at this question that states determine the exact value of the following expression without the help of the calculator where the expression is C can of 19 pi divided by three. Now the answer choices provided are a two B one half C square root of two divided by two and D negative two. Now for a question like this where we have uh an angle that is greater than the range that we want. So in a coordinate system, trigonometric functions fall within a range. Now let's say we have the coordinate system, we have a Y axis and an X axis. Now we start at zero on the right side of the X axis. The top half of the y axis will have pi divided by two. The left side of the X axis will be pi the bottom half of the Y axis will be three pi divided by two. And we once we go back to zero, that will be, it will be zero or two pi so that will be the full revolution. And we move counterclockwise meaning in a positive direction. Now, once you do that full revolution of two pie, since you are working with a circle, every value would just repeat, doesn't matter how many times you go around that circle. So our range is zero is less than theta is less than two pi that's the range that we're working with. Now, 19 pi divided by three is out of that range. So we have to think about how can we fall within this range? So we have C can of 19 pie divided by three. Now that can be rewritten as C cans of. Now think about you start at zero, you do one revolution that is two pi another revolution is four P and so on. Now we are, we can go around six pie plus pi divided by three. Now, why is this? Well, six pi is the same thing as saying 18 pi divided by three. And then we add that final pi divided by three to get to the 19 pi divided by three that we want six pi is the same as going around three times because as we mentioned, one revolution is two pi. So six pi is going around three times. So we go around three times, we can ignore those three times because all we care about is that final revolution that we don't complete. Where do we end up? And since we said we have six pi plus pi divided by three, we end up at pi divided by three. Therefore, she can't of 19 pie divided by three is equal to c cans of I divided by three. They fall in the same location in the circle just in a different revolution. So since we have this information, now we can turn to the unit circle and use it to our advantage. So recall from unit circle and these are values that we should be able to recall right away. So let's say pi divided by three, which is where we land there is a point at that angle. That point has a X component of one half, comma a Y component of square root of three divided by two. Now, something else we should know in the unit circle is that the X is associated with cosine of theta and the Y is associated with sine of theta. So if we have C cans of pi divided by three C can is the reciprocal of cosine. So C can of pi divided by three is equal to one divided by cosine of pi divided by three. And we just said that cosine is associated with the X component. So we'll have one divided by the X component at pi divided by three, which is one half. And if we have one divided by one half, that is equal to two. So we have second of 19 pie divided by three equals seconds of pi divided by three which equals two. And just like that we've completely simplified this problem to figure out our answer and match it up with answer choice. A and we can see how important it is to be able to recall different values from the unit circle as well as rewrite our equations to fit in with the range that we want. 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. sec 22𝜋 3

Key Concepts

Understanding the Secant Function

Evaluating Trigonometric Functions at Special Angles

Using the Unit Circle for Exact Values

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