Trigonometric Functions: The Unit Circle

1. Trigonometric Functions: The Unit Circle Section 4.2

2. Objectives • Find a point on the unit circle given one coordinate and the quadrant in which the point lies. • Determine the coordinates of a point on the unit circle given a point on the unit circle. • State the sign of the sine or cosine value of an angle based on the quadrant in which the terminal side of an angle occurs. • State the sine and cosine values of an angle (measured in radians) where the angles have a measure of

3. Objectives • Determine the tangent, cotangent, secant, and cosecant values of an angle given a point on the unit circle. • State the sign of the tangent, cotangent, secant, and cosecant value of an angle based on the quadrant in which the terminal side of an angle occurs. • Determine the tangent, cotangent, secant, and cosecant values of an angle (measured in radians) where the angles have a measure of

4. Vocabulary • quadrant • sine of an angle • cosine of an angle • terminal side of an angle • initial side of an angle • tangent of an angle • cotangent of an angle • secant of an angle • cosecant of an angle

5. Unit Circle Each point on the unit circles has an x-coordinate and a y-coordinate. Any angle, θ, in standard position (vertex at the origin and initial side on the positive x-axis) will have a terminal side that intersects the unit circle in a single point. We define the sin(θ) as the value of the y-coordinate of the point of intersection between the terminal side of the angle and the unit circle. We define the cos(θ) as the value of the x-coordinate of the point of intersection between the terminal side of the angle and the unit circle. (cos(θ), sin(θ))=(x, y) θ

6. If the point is on the unit circle in quadrant IV, then find y. The coordinates of all points on the unit circle satisfy the equation This means that all we have to do is plug in the value of the x-coordinate into the equation and solve for y. continued on next slide

7. If the point is on the unit circle in quadrant IV, then find y. continued on next slide

8. If the point is on the unit circle in quadrant IV, then find y. Since there can only be one answer to this question, we have to determine if the answer is the positive number or the negative number. We are told that the point is in quadrant IV. In quadrant IV, the x-values are positive and the y-values are negative. This means that the correct answer for our problem is

9. If P(t) has coordinates (0.141, 0.99), find the coordinates of each point indicated below. The first thing that we want to do for this problem put the point P(t) on the picture. P(t)=(0.141, 0.99) t t + π In order to answer part 1, it will help to draw the angle t + π on the picture. Keep in mind that the distance π is half way around the circle. continued on next slide

10. If P(t) has coordinates (0.141, 0.99), find the coordinates of each point indicated below. Because of the symmetries of the circle, the new point is the same horizontal distance away from the from the y-axis as the original point. It is just in the opposite direction. Similarly, the new point is the same vertical distance from the x-axis as the original point. It is just in the opposite direction. This means that the only thing that changes to the coordinates as we go from the old point to the new point is the sign of each coordinate. Thus P(t)=(0.141, 0.99) t t + π continued on next slide

11. If P(t) has coordinates (0.141, 0.99), find the coordinates of each point indicated below. In order to answer part 2, it will help to draw the angle -t on the picture. Keep in mind that the angle –t is the same distance as the angle t only in the clockwise direction. P(t)=(0.141, 0.99) t Because of the symmetries of the circle, the new point is the same horizontal distance away from the from the y-axis as the original point. This time it is in the same direction. -t continued on next slide

12. If P(t) has coordinates (0.141, 0.99), find the coordinates of each point indicated below. Similarly, the new point is the same vertical distance from the x-axis as the original point. It is just in the opposite direction. This means that the only thing that changes to the coordinates as we go from the old point to the new point is the sign of the y-coordinate. Thus P(t)=(0.141, 0.99) t -t continued on next slide

13. If P(t) has coordinates (0.141, 0.99), find the coordinates of each point indicated below.- In order to answer part 3, it will help to draw the angle t - π on the picture. Keep in mind that the distance π is half way around the circle. In this case since it is – π we will be going clockwise half way around the circle P(t)=(0.141, 0.99) t t-π Notice that since the beginning of the angle is t (which is positive), we start off in the counter-clockwise direction. Once we get to t, we turn and go back in the clockwise direction a distance of π. continued on next slide

14. If P(t) has coordinates (0.141, 0.99), find the coordinates of each point indicated below.- Because of the symmetries of the circle, the new point is the same horizontal distance away from the from the y-axis as the original point. It is just in the opposite direction. Similarly, the new point is the same vertical distance from the x-axis as the original point. It is just in the opposite direction. This means that the only thing that changes to the coordinates as we go from the old point to the new point is the sign of each coordinate. Thus P(t)=(0.141, 0.99) t t-π continued on next slide

15. If P(t) has coordinates (0.141, 0.99), find the coordinates of each point indicated below. In order to answer part 4, it will help to draw the angle -t - π on the picture. Keep in mind that the distance π is half way around the circle. This time we start in the clockwise direction a distance of t and then continue in the clockwise direction half way around the circle further. P(t)=(0.141, 0.99) t -t - π continued on next slide

16. If P(t) has coordinates (0.141, 0.99), find the coordinates of each point indicated below. Because of the symmetries of the circle, the new point is the same horizontal distance away from the from the y-axis as the original point. It is just in the opposite direction. Similarly, the new point is the same vertical distance from the x-axis as the original point. This however is in the same direction. This means that the only thing that changes to the coordinates as we go from the old point to the new point is the sign of the x-coordinate. Thus P(t)=(0.141, 0.99) t -t - π

17. Find the terminal point P(x, y) on the unit circle determined by the value of point of intersection Since our angle is the one that has a terminal side that runs along the positive y-axis, this terminal side of the angle intersects the unit circle at (0, 1). This means that the x-coordinate of the point is 0 and the y-coordinate of the point is -1.

18. Sine and Cosines for Basic Quadrant I Angles For the next few slides we are going to find the coordinates of the points where the terminal side of three standard angles intersect with the unit circle. This will allow us to define the sine and cosine values for the following quadrant I angles:

19. If , find the sin(t) and cos(t). For this problem, we are going to draw a right triangle the length of whose legs are equal to the coordinates of the point where the angle intersects the unit circle. The dark blue triangle that we have drawn has two non-right angles in it. One of the angles is π/4 (or 45 degrees). Since the sum of the angles of a triangle add up to 180 degrees, the other non-right angle must have a measure of 45 degrees also. continued on next slide

20. y x If , find the sin(t) and cos(t). This makes the dark blue triangle an isosceles triangle. In an isosceles triangle the lengths of the sides opposite the angles that are equal are also equal. This means that the sides marked x and y are equal in length. Now a right satisfies the Pythagorean Theorem. This means The hypotenuse of the triangle is equal to 1 since it is the radius of a unit circle. continued on next slide

21. y x If , find the sin(t) and cos(t). Since both x and y are the same length, we can replace y with x (we could replace x with y) and end up at the same place. Once we do this we can solve the equation for x. continued on next slide

22. y x If , find the sin(t) and cos(t). Since x is a length, x must be positive. Thus Notice that x has the denominator rationalized. Since x and y are the same, then Thus and

23. If , find the sin(t) and cos(t). Our aim here is to get a right triangle whose legs have the same measure as the coordinates of the point where the terminal side of the angle π/3 (60 degrees) intersects the unit circle. This process will take a little more work than with the 45 degree angle. We will start by completing a triangle by drawing the red line connecting the points where the initial side and terminal side of the angle intersect the unit circle. continued on next slide

24. If , find the sin(t) and cos(t). Now we will put some labels on our figure. Side U and side V are both of length 1 since they are both radii of a unit circle. This means that our triangle is an isosceles triangle. In an isosceles triangle the angles opposite the equal sides are equal. Thus the marked angles are equal. We will say they measure a. We now need to find the measure of each of those angles. The three angle of the triangle add up to 180. We know one of the angles is 60. Thus we can calculate a. U W a a V continued on next slide

25. U W a a V If , find the sin(t) and cos(t). To find a we have Since all of the angles of our triangle are the same measure, we have an equiangular (also equilateral) triangle. continued on next slide

26. U W a a V If , find the sin(t) and cos(t). An altitude of a triangle is a line that goes from one vertex and is perpendicular to the opposite side (or an extension of the opposite side. We are going to draw an altitude in our triangle from the point where the terminal side of our angle intersect the unit circle to the x-axis. In an equilateral triangle, altitudes have the property that they bisect (cut into two equal pieces) both the angle from which they original and the opposite side. This means that the distance from the origin to the intersection of the altitude and the x-axis is ½ because the length of the original side was 1. continued on next slide

27. U W a y a V ½ If , find the sin(t) and cos(t). We can now draw a right triangle. The length of the leg along the positive x-axis is ½. This means that the x-coordinate of our point is ½. The length of the hypotenuse of the triangle is 1 since it is a radius of a unit circle. Using this and the Pythagorean Theorem, we can calculate y. continued on next slide

28. U W a y a V ½ If , find the sin(t) and cos(t). Since y represents a length, it must be positive. Thus Thus and

29. If , find the sin(t) and cos(t). The same process that was used for π/3 will work for π/6. The only difference is that we will make the first triangle by connecting the point where the terminal side of the angle intersections the unit circle and where the y-axis intersects the unit circle. We know that the angle in this triangle whose vertex is the origin is 60 degrees since it is complementary to the π/6 (30 degree) angle that we are given. continued on next slide

30. If , find the sin(t) and cos(t). The process then continues with the altitude intersecting the y-axis (instead of the x-axis as in the previous example). Thus and

31. Other Trigonometric Functions Once we have defined the sine and the cosine for an angle, we can define all of the other trigonometric functions in terms of the sine and cosine functions.

32. Quotient Identities

33. Reciprocal Identites

34. Pythagorean Identity