Lesson Video: Trigonometric Ratios on the Unit Circle | Nagwa Lesson Video: Trigonometric Ratios on the Unit Circle | Nagwa

# Lesson Video: Trigonometric Ratios on the Unit Circle

In this video, we will learn how to relate the x- and y-coordinates of points on the unit circle to trigonometric functions.

15:41

### Video Transcript

In this video, we will learn how to use the fact that the quadrant where an angle lies determines the sign of its trigonometric functions sine, cosine, and tangent to solve equations. Letβs start by thinking about what we know about angles on a coordinate plane.

On a coordinate plane, the center is the origin. For standard angles measured in the coordinate plane, the positive π₯-axis will be the initial side, the starting point of our measurement. This positive π₯-axis and the beginning represents zero degrees. By the time we get to the positive π¦-axis, weβve gone 90 degrees. The negative π₯-axis will be 180 degrees. The negative π¦-axis will be 270 degrees. And all the way back around for a full turn is 360 degrees. In radians, that would be zero, π over two, π, three π over two, and then two π.

The ray that represents where we stop measuring our angle is called the terminal side. And the angle and standard position will be located between the initial and terminal side. But we want to specifically consider angles that fall on the unit circle. The unit circle is centered at the origin and has a radius of one unit. For some coordinate π₯, π¦ in the first quadrant, the lengths π₯ and π¦ become the legs of a right triangle. And because we know that the radius of this circle is one, the hypotenuse of this right triangle will be one.

If we label the side lengths here π₯ and π¦, we can take the Pythagorean theorem, which tells us π squared plus π squared equals π squared, in the unit circle π₯ squared plus π¦ squared equals one. We can also say that the equation of the unit circle must be π₯ squared plus π¦ squared equals one. But what we wanna do now is consider a bit of trigonometry with regards to the unit circle sine, cosine, and tangent. We know the sine of an angle represents the opposite side length over the hypotenuse, the cosine of an angle π represents the adjacent side length over the hypotenuse, and the tangent relationship of an angle is the opposite side length over the adjacent side length.

For our angle π inside the unit circle, the opposite side length will always be the π¦-value, itβs the vertical leg. The sin of π is then π¦ over one since in the unit circle the hypotenuse is always one. sin of π is, therefore, equal to π¦ in the unit circle. Similarly, the adjacent side will be the horizontal leg, the π₯-value. And that means in the unit circle, the cos of π will be equal to π₯. And this means that the tan of angle π will be equal to π¦ over π₯.

Before we look at some examples, thereβs one final thing we need to consider. And that is the sine of angles in quadrants. If we keep our unit circle, and we look at our π₯, π¦ in the first quadrant, in quadrant one, all π₯-values are positive and all π¦-values are positive. This means that the sin of π would be positive since π¦ is positive, the cos of π would be positive since π₯ is positive, and since π₯ and π¦ are both positive, the tan of π will also be positive. This means that in quadrant one, all trig relationships are positive.

But as we move to quadrant two and create a right angle from the point negative π₯, π¦, weβre dealing with negative π₯-values and positive π¦-values. The π¦-value is positive in the second quadrant, making the sine value positive. But the π₯-value is negative. And that means in quadrant two, the cosine will be negative. With a positive value and a negative value, The tangent then becomes negative. So, we can generalize and say in quadrant two, the sine is positive but the cosine and tangent will be negative.

Moving around to quadrant three, in quadrant three, the π₯-value and the π¦-value will be negative. This means the sine will be negative, the cosine will be negative. However, with tangent, negative π¦ over negative π₯ will be positive. And so, in quadrant three, sine and cosine are negative, while tangent is positive. And finally, in quadrant four, the π₯-values will be positive and the π¦-values will be negative. This gives us a negative sine value, a positive cosine value, and a negative tangent value.

One way that you can remember these sine values is by using the CAST diagram that looks like this. The CAST diagram indicates which trig value will be positive in which quadrant. In quadrant one, all are positive. In quadrant two, sine is positive. In quadrant three, tangent is positive. And in quadrant four, cosine is positive. Now, weβre ready to consider some examples.

Find sin of π, given π is in standard position and its terminal side passes through the point three-fifths, negative four-fifths.

Itβs probably helpful here to sketch a coordinate plane. On this coordinate plane, we want to plot the point three-fifths, negative four-fifths, which is here. If we know that the terminal side of our angle goes through this point and that our angle is in standard position, then the initial side will be the ray that begins at the origin and goes through the positive π₯-axis. That means the angle weβre interested in is the space from the initial side to the terminal side. However, to calculate this, weβll form a right angle with the π₯-axis.

When we do that, we end up with a right-angled triangle. And weβll use the angle created with the π₯-axis to solve. We can solve this using right-triangle trigonometry, where we have a right triangle with side lengths three-fifths and four-fifths. We remember that sin of π will be the opposite over the hypotenuse. However, we donβt currently know the hypotenuse. We donβt know the distance from the origin to the point three-fifths, negative four-fifths. We can use the Pythagorean theorem to find that, which means three-fifths squared plus four-fifths squared equals π squared. Nine twenty-fifths plus sixteen twenty-fifths equals π squared. 25 over 25 equals one. And if π squared equals one, then π must be equal to one.

Now that we know the hypotenuse is equal to one, we can say something else about this angle in our coordinate plane. And that is that our point falls on the unit circle. The unit circle has its center at the origin and has a radius of one. Weβre looking for the sin of π. Itβs equal to the opposite over the hypotenuse. Also in a unit circle, the sin of π is equal to its π¦-coordinate. The π¦-coordinate for this point is negative four-fifths. And so, we say that the sin of π equals negative four-fifths. And if we wanted to check this, we could remember based on our CAST diagram that for angles falling in the fourth quadrant, the cosine is positive, but the sine and tangent will be negative, which confirms that the sin of our angle π is equal to negative four-fifths.

Letβs consider another example.

Suppose π is a point on the unit circle corresponding to the angle four π over three. Is there another point on the unit circle representing an angle in the interval zero to two π that has the same tangent value? If yes, give the angle.

First, we might wanna sketch a coordinate plane and then add a unit circle, which is a circle with the center at the origin and a radius of one. From there, we might also want to label our coordinate plane with radians beginning at zero, π over two, π, three π over two, and two π. Because we have the interval zero to two π, we know weβre only interested in one full turn. Our point π is on the unit circle and corresponds to the angle four π over three. This means our first job is to find out where the angle four π over three would land.

I know for π over three is greater than π, but itβs probably worth comparing to find out if four π over three is greater than or less than three π over two. If we give these fractions common denominators, four π over two becomes eight π over six and three π over two becomes nine π over six. Since eight π over six is less than nine π over six, we can say that four π over three is less than three π over two. And that means point π is going to fall in our third quadrant and that four π over three would be this angle.

Since we know that our angle falls in the third quadrant, we can use the CAST diagram, which will tell us that the tangent of the angle in the third quadrant is going to be positive. In order for us to find another point inside the unit circle that has the same tangent value, weβll be looking for the other place where the tangent value could be positive. And that will be in the first quadrant. In the first quadrant, all trig values will be positive.

But in order for us to find what the value of that angle would be in the first quadrant, we need to break up our angle four π over three into smaller parts. We could say that four π over three is equal to π plus π over three, the distance from zero to π and then an additional π over three. The right triangle created inside the unit circle four π over three in the third quadrant would look like this.

And in the first quadrant, there would be some point such that we would be dealing with the angle of π over three. In the first quadrant, that would have π₯, π¦ coordinates. And in the third quadrant, it would have negative π₯, negative π¦ coordinates. And we know that in a unit circle, the tan of π will be equal to π¦ over π₯. And we would say that the tan of four-thirds π is equal to negative π¦ over negative π₯ and the tan of π over three is equal to π¦ over π₯. But we simplify negative π¦ over negative π₯ to just π¦ π₯. And so, weβve shown that yes, there is another angle in this interval that has the same tangent value. And itβs the angle π over three.

In our next example, weβll consider an application of the unit circle.

Consider a windmill with the blades of length one meter. The position of the top π of a given blade is given by the coordinates π, π, which depends upon the angle π as shown. Express π and π as functions of the measure of angle π in radians. If angle π at a certain time is five-thirds π, what will it be after the blade has completed half a rotation?

Because the length of the blades are one meter and in our diagram, this represents the radius of the windmill, we can use our knowledge of the unit circle to help us solve this problem. The distance from the center of the windmill to point π would be one. And weβre told that the point π is located at π, π. And so, we can create a right triangle with the π₯-axis and say that its side lengths are π and π, respectively. We use the distance from the terminal side to the π₯-axis to calculate the angle π. We know that in the unit circle, we can represent the angle as a sine and cosine relationship. For our angle π, the opposite side length would be π and its hypotenuse is one meter, so we have sin of π equals π over one. And we can represent π as sin of π.

Similarly, if we look at cosine, we end up with the side length π over one, which means we can say that the distance π must be equal to the cos of the angle π. And without more information, this is as far as we can go with these two functions. We can say that π equals sin of π and π equals cos of π.

Part two of our questions says if the angle of π at a certain time is five π over three, what will it be after the blade has completed a half a rotation? First of all, weβve already been told that weβre operating in radians, and so it might be helpful to label our coordinate plane. Beginning at the π₯-axis, zero radians, then π over two radians, π radians, three π over two radians, and a full turn, which is two π radians. In a system like this, a full turn is two π. And therefore, a half turn is going to be equal to π radians.

If we start with the angle π five π over three radians and we add a half a rotation, weβre adding π to that angle. And just like adding any fractions, we need a common denominator. We can write π as three π over three. We add five π plus three π to get eight π. And the denominator wouldnβt change. The angle π after a half rotation would be eight π over three.

Before we finish, letβs review our key points. The unit circle is a circle with the center at the origin and a radius of one unit. For any point on the unit circle π₯, π¦, the sin of the angle π created by that point will be equal to π¦ and the cos of angle π will be equal to π₯. Additionally, π₯ squared plus π¦ squared must equal one. We also saw that the CAST diagram can be used to identify the signs of trigonometric functions in each quadrant.

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