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.

[π, 3π/2] ; tan s = √3

Accepted Answer

Welcome back. I am so glad you're here. We're asked to find the exact value of P. The angle P lies in the interval pi divided by two to pi. And we're told that the tangent of P equals negative square root three. Our answer choices are answer choice A four pi divided by three answer choice B five pi divided by three answer choice C five pi divided by six and answer choice D two pi divided by three. All right. So we're looking for the exact value of this angle P and it's equal to the tangent of P is equal to negative square root three, which is kind of familiar. So what we can do is take a look at the unit circle and see if it looks like it might be on there. 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 over that we have our circle. Every point on the circle is exactly one unit away from the origin. That is not true of what I drew. But you can imagine that it is. Now I am only going to draw a part of this for the entire unit circle. Definitely check out the image of the unit circle. In your textbook, we're told that our angle P lies in the interval from pi divided by two to pi well pi divided by two is the positive part of the y axis and pi is the negative part of the X axis. So it's in between those two, we're talking about quadrant two, I'll draw three of the angles on the unit circle that are in quadrant two. The first one closer to the positive part of the Y axis than the negative part of the X axis is two pi divided by three and its XY coordinates are negative one half square root three divided by two. The next angle directly in between the positive part of the Y axis and the negative part of the X axis is three pi divided by four. Its xy coordinates are negative square root two divided by two square root two, divided by two. And the last one I'll draw is closer to the negative part of the X axis than the positive part of the Y axis. It is five pi divided by six and its xy coordinates are negative square root three divided by two, one half. And so why is this important? We were looking for something like negative square root three. We don't have anything that's exactly like that. But we do recall from previous lessons that the tangent of an angle on the unit circle, we'll call that S is equal to Y divided by X. So can we take any of these Y values, divide them by our X values and end up with negative square three? Let's see if that's true. So we're going to end up with a negative square root three. So it would be helpful if our Y value in the numerator had a square root three in its numerator. Well, that would be for our two pi divided by three. The Y value there is square root three divided by two. So if we take that and divide it by the X value, which is negative one half. Does that equal negative square root three, let's find out we have fractions within a fraction. So I'm going to rewrite this as square root three divided by two divided by, with a division symbol negative one half because that reminds me to multiply by the second fractions, reciprocal square root three divided by two, multiplied by negative two. Well, those twos cancel out. So we have square root three multiplied by negative one which is negative square root three. Yes, it is true here. So our Y divided by X is negative square root three, which means that our angle is this two pi divided by three. So P equals two pi divided by three. We look at our answer choices and this matches with answer choice. D 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.

[π, 3π/2] ; tan s = √3

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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.[π, 3π/2] ; tan s = √3

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.

[π, 3π/2] ; tan s = √3