Lesson Video: Trigonometric Ratios on the Unit Circle Mathematics • 10th Grade

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

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Video Transcript

In this video, we will learn how to use the fact that the quadrant where an angle lies determines the signs of its sine, cosine, and tangent and solve trigonometric equations. We will begin by considering the unit circle.

The unit circle is a circle with a radius of one whose center lies at the origin of a coordinate plane. For any point π‘₯, 𝑦 on the unit circle, a right triangle can be formed as in the diagram. The hypotenuse of this right triangle makes an angle πœƒ with the positive π‘₯-axis. Using right triangle trigonometry, we can define the trigonometric functions in terms of the unit circle. Recall sin πœƒ equals the opposite over the hypotenuse, cos πœƒ equals the adjacent over the hypotenuse, and tan πœƒ equals the opposite over the adjacent.

Since the opposite side has length 𝑦, the adjacent side has length π‘₯, and the hypotenuse has length one, we have the following equations. So 𝑦 is equal to sin πœƒ, π‘₯ is equal to cos πœƒ, and 𝑦 over π‘₯ is equal to tan πœƒ. We note that tan πœƒ is not defined when π‘₯ equals zero. We also observe that, while we have derived these definitions for an angle πœƒ in quadrant I, they hold for an angle in any quadrant. Summarizing, the π‘₯- and 𝑦-coordinates of a point on the unit circle given by an angle πœƒ are defined by π‘₯ equals cos πœƒ and 𝑦 equals sin πœƒ.

In our first example, we will demonstrate how we can use these definitions of trigonometric functions in the unit circle to find exact values, given information about the terminal side of an angle.

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

An angle is said to be in standard position if the vertex is at the origin and the initial side lies on the positive π‘₯-axis. The angle is measured in a counterclockwise direction from the initial side to the terminal side. Hence, the angle πœƒ is as shown. We draw a right triangle with side lengths of three-fifths units and four-fifths units.

Now we can use the Pythagorean theorem to calculate the value of the missing dimension in the triangle. Three-fifths squared plus four-fifths squared is equal to 𝑐 squared. The left-hand side simplifies to nine over 25 plus 16 over 25. So 𝑐 squared is equal to 25 over 25, which equals one. Square rooting both sides, and since 𝑐 must be positive, we have 𝑐 is equal to one. This tells us that the point three-fifths, negative four-fifths lies on the unit circle.

We recall that the π‘₯- and 𝑦-coordinates of a point on the unit circle given by an angle πœƒ are defined by π‘₯ equals cos πœƒ and 𝑦 equals sin πœƒ. Sin πœƒ is therefore equal to the value of the 𝑦-coordinate of the point, which is negative four-fifths. The correct answer is sin πœƒ equals negative four-fifths.

In addition to the standard trigonometric functions, it is also possible to define the reciprocal trigonometric functions. The reciprocal of a number π‘₯ is one over π‘₯. For an angle πœƒ, where πœƒ is a real number, the reciprocal trigonometric functions are as follows. csc πœƒ equals one over sin πœƒ, sec πœƒ equals one over cos πœƒ, and cot πœƒ equals one over tan πœƒ, where sin πœƒ, cos πœƒ, and tan πœƒ are not equal to zero.

Since we can write the standard trigonometric functions in terms of the unit circle, it is also possible for us to write the reciprocal functions in terms of the unit circle. That is, let us once more consider a point π‘₯, 𝑦 on the unit circle with angle πœƒ to the positive π‘₯-axis, where π‘₯ is equal to cos πœƒ and 𝑦 equals sin πœƒ. Then the reciprocal trigonometric functions can be written as follows. csc πœƒ equals one over 𝑦, sec πœƒ equals one over π‘₯, and cot πœƒ equals π‘₯ over 𝑦, where the denominators cannot equal zero.

Let us consider an example where we can use the unit circle to find the exact value of the secant function.

Find sec πœƒ, given that πœƒ is in standard position and its terminal side passes through the point four-fifths, three-fifths.

An angle is said to be in standard position if the vertex is at the origin and the initial side lies on the positive π‘₯-axis. The angle is measured in a counterclockwise direction from the initial side to the terminal side. Hence, the angle πœƒ is as shown.

To calculate the value of sec πœƒ, we will begin by determining whether the point with coordinates four-fifths, three-fifths lies on the unit circle. To do this, we can consider a right triangle with side lengths of four-fifths units and three-fifths units as shown. Then we can calculate the length of the hypotenuse, 𝑐, by using the Pythagorean theorem.

We have four-fifths squared plus three-fifths squared is equal to 𝑐 squared. This simplifies to 16 over 25 plus nine over 25 equals 𝑐 squared. So 𝑐 squared is equal to 25 over 25, which equals one. Square rooting both sides, and since 𝑐 must be positive, we have 𝑐 is equal to one. So we have shown that the point four-fifths, three-fifths lies on the unit circle.

The π‘₯-and 𝑦-coordinates of a point on the unit circle given by an angle πœƒ are defined by π‘₯ equals cos πœƒ and 𝑦 equals sin πœƒ. We recall that sec πœƒ is equal to one over cos πœƒ. So sec πœƒ is equal to one over π‘₯. Substituting in the π‘₯-coordinate, we have sec πœƒ equals one divided by four-fifths, which is equal to five over four, or five-quarters. If πœƒ is in standard position and the terminal side of the angle passes through the point four-fifths, three-fifths, then sec πœƒ is equal to five-quarters.

We will now consider one final example where we will demonstrate how to use the unit circle to evaluate a simple trigonometric function.

The terminal side of angle 𝐴𝑂𝐡 in standard position intersects the unit circle 𝑂 at the point 𝐡 with coordinates three over root 10, 𝑦, where 𝑦 is greater than zero. Find the value of sin 𝐴𝑂𝐡.

An angle is said to be in standard position if the vertex is at the origin and the initial side lies on the positive π‘₯-axis. Since angle 𝐴𝑂𝐡 is in standard position and 𝐡 is not on the π‘₯-axis, the point 𝐴 must lie on the positive π‘₯-axis. We can therefore sketch angle 𝐴𝑂𝐡 equals πœƒ on the unit circle. Since both the π‘₯- and 𝑦-coordinates are positive, point 𝐡 lies in the first quadrant.

We know that the π‘₯- and 𝑦-coordinates of a point on the unit circle given by an angle πœƒ are defined by π‘₯ equals cos πœƒ and 𝑦 equals sin πœƒ. The value of sin 𝐴𝑂𝐡 is therefore equal to the value of the 𝑦-coordinate of point 𝐡.

By representing triangle 𝐴𝑂𝐡 as a right triangle, we can find the value of 𝑦 by using the Pythagorean theorem. We have 𝑦 squared plus three over root 10 squared is equal to one squared. Simplifying this, our equation becomes 𝑦 squared plus nine over 10 equals one. Subtracting nine-tenths from both sides, we have 𝑦 squared is equal to one-tenth. Square rooting both sides, and since 𝑦 is greater than zero, we obtain 𝑦 is equal to one over root 10 units. The value of sin 𝐴𝑂𝐡 is one over root 10.

We will finish this video by recapping some of the key points.

The unit circle is a circle of radius one, centered at the origin of the coordinate plane. The π‘₯- and 𝑦-coordinates of a point on the unit circle given by an angle πœƒ are defined by π‘₯ equals cos πœƒ and 𝑦 equals sin πœƒ. The tangent ratio can also be defined for points on the unit circle π‘₯, 𝑦, where π‘₯ is nonzero, such that tangent πœƒ is equal to 𝑦 over π‘₯. The reciprocal trigonometric functions are defined as follows. csc πœƒ is equal to one over 𝑦, sec πœƒ is equal to one over π‘₯, and cot πœƒ is equal to π‘₯ over 𝑦.

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