Find the exact value of s in the given interval that has the g...

Question

Find the exact value of s in the given interval that has the given circular function value.

[3π/2, 2π] ; tan s = -1

Accepted Answer

Welcome back. I am so glad you're here. We're asked to find the exact value of P and the angle P lies in the interval from pi to three pi divided by two. We're told that the tangent of P equals the square root of three divided by three. Our answer choices are answer choice. A pi divided by six answer choice B seven pi divided by six answer choice C four pi divided by three and answer choice D five pi divided by six. All right. So we're looking for an angle measure P and the tangent of it equals the square root three divided by three. And that's pretty familiar what we can do is take a look at the unit circle and see if there's an exact value on there. So I'm going to draw a rough sketch of the unit circle. We have a vertical y axis and a horizontal x axis. They come together at the origin in the middle. The unit circle lies over that and every point on the unit circle is exactly one unit away from the origin. That is not true of what I drew. But you can use your imagination and we're told that the angle we're looking for P lies in the interval from pi to three pi divided by two. Well, pi is the negative part of the X axis and three pi divided by two is the negative part of the y axis. So we're talking about quadrant three. Now, for all of the details of the unit circle, definitely check out the figure of the unit circle. In your textbook. I'm only going to draw the parts the angles that are in quadrant three. So closer to the negative part of the X axis than the negative part of the Y axis, we have the angle seven pi divided by six. Its xy coordinates are negative square root three divided by two negative one half. Next directly in between the negative parts of the X and Y axis. We have the angle five pi divided by four. It's xy coordinates are negative square root two divided by two and negative square root two divided by two. And then closer to the negative part of the Y axis than the negative part of the X axis. We have the angle four pi divided by three. It's xy coordinates are negative one half, negative square root three divided by two and so none of those are exactly square root three divided by three. But we're very close and we recall from previous lessons that the tangent of an angle on the unit circle will call that S is equal to Y divided by X. And so if we take any of these Y values and divide them by our X values, can that give us square root three divided by three? Well, maybe, let's see, we want one that has a square root three in it somewhere. So we can rule out the five pi divided by four. That's just a bunch of square at twos and twos. So it could be seven pi divided by six or four pi divided by three. Well, if we look at four pi divided by three, that has a square root three in the numerator of the Y value. But all that's left are ones and twos. And if you put that in your calculator or you do the math, that's not going to end up being square root three divided by three. But if you take the seven pi divided by six Y value negative one half and divide that by its X value negative square root three divided by two. Well, dividing fractions is the same as multiplying by the reciprocal of the second fraction. So we can write negative one half multiplied by two divided by the square root of three. And that's negative. I'll bring that negative up to the numerator multiplying straight across one multiplied by two is a positive two. Now divided by two, multiplied by square root of three is two square root three. So that's not exactly square root three divided by three. But we want to rationalize this. So we'll multiply both the numerator and the denominator by the square root of three. Now, in the numerator, we have two square root three divided by, in the denominator, it's two multiplied by square root of three, multiplied by square root of three, which is just three. But we noticed that these two factors cancel out. So this simplifies to be the square root of three divided by three that we're looking for. So that means that the angle we're looking for is this seven pi divided by six because dividing its Y value by its X value gets us that square root three divided by three. So P here, our angle is seven pi divided by six. We look at our answer choices and this matches with answer choice. B well done. We'll catch you on the next one.

Struggling with Trigonometry?

Find the exact value of s in the given interval that has the given circular function value.

[3π/2, 2π] ; tan s = -1

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Defining the Unit Circle practice set

Struggling with Trigonometry?

Find the exact value of s in the given interval that has the given circular function value.[3π/2, 2π] ; tan s = -1

Watch next

Defining the Unit Circle practice set

Find the exact value of s in the given interval that has the given circular function value.

[3π/2, 2π] ; tan s = -1