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10.2 The Unit Circle: Cosine and Sine 717
10.2 The Unit Circle: Cosine and Sine
In Section 10.1.1, we introduced circular motion and derived a formula which describes the linear velocity of an object moving on a circular path at a constant angular velocity. One of the goals of this section is describe the position of such an object. To that end, consider an angle θ in standard position and let P denote the point where the terminal side of θ intersects the Unit Circle. By associating the point P with the angle θ, we are assigning a position on the Unit Circle to the angle θ. The x-coordinate of P is called the cosine of θ, written cos(θ), while the y-coordinate of P is called the sine of θ, written sin(θ).^1 The reader is encouraged to verify that these rules used to match an angle with its cosine and sine do, in fact, satisfy the definition of a function. That is, for each angle θ, there is only one associated value of cos(θ) and only one associated value of sin(θ).
x
y
1
1
θ
x
y
1
1 P (cos(θ), sin(θ))
θ
Example 10.2.1. Find the cosine and sine of the following angles.
- θ = 270◦^ 2. θ = −π 3. θ = 45◦^ 4. θ = π 6 5. θ = 60◦
Solution.
- To find cos (270◦) and sin (270◦), we plot the angle θ = 270◦^ in standard position and find the point on the terminal side of θ which lies on the Unit Circle. Since 270◦^ represents 34 of a counter-clockwise revolution, the terminal side of θ lies along the negative y-axis. Hence, the point we seek is (0, −1) so that cos
( (^3) π 2
= 0 and sin
( (^3) π 2
- The angle θ = −π represents one half of a clockwise revolution so its terminal side lies on the negative x-axis. The point on the Unit Circle that lies on the negative x-axis is (− 1 , 0) which means cos(−π) = −1 and sin(−π) = 0. (^1) The etymology of the name ‘sine’ is quite colorful, and the interested reader is invited to research it; the ‘co’ in ‘cosine’ is explained in Section 10.4.
718 Foundations of Trigonometry
x
y
1
1
P (0, −1)
θ = 270◦
Finding cos (270◦) and sin (270◦)
x
y
1
1
P (− 1 , 0)
θ = −π
Finding cos (−π) and sin (−π)
- When we sketch θ = 45◦^ in standard position, we see that its terminal does not lie along any of the coordinate axes which makes our job of finding the cosine and sine values a bit more difficult. Let P (x, y) denote the point on the terminal side of θ which lies on the Unit Circle. By definition, x = cos (45◦) and y = sin (45◦). If we drop a perpendicular line segment from P to the x-axis, we obtain a 45◦^ − 45 ◦^ − 90 ◦^ right triangle whose legs have lengths x and y units. From Geometry,^2 we get y = x. Since P (x, y) lies on the Unit Circle, we have x^2 + y^2 = 1. Substituting y = x into this equation yields 2x^2 = 1, or x = ±
1 2 =^ ±
√ 2
Since P (x, y) lies in the first quadrant, x > 0, so x = cos (45◦) =
√ 2 2 and with^ y^ =^ x^ we have y = sin (45◦) =
√ 2
x
y
1
1
P (x, y)
θ = 45◦
θ = 45◦
x
y
P (x, y)
(^2) Can you show this?
720 Foundations of Trigonometry
In Example 10.2.1, it was quite easy to find the cosine and sine of the quadrantal angles, but for non-quadrantal angles, the task was much more involved. In these latter cases, we made good use of the fact that the point P (x, y) = (cos(θ), sin(θ)) lies on the Unit Circle, x^2 + y^2 = 1. If we substitute x = cos(θ) and y = sin(θ) into x^2 + y^2 = 1, we get (cos(θ))^2 + (sin(θ))^2 = 1. An unfortunate^4 convention, which the authors are compelled to perpetuate, is to write (cos(θ))^2 as cos^2 (θ) and (sin(θ))^2 as sin^2 (θ). Rewriting the identity using this convention results in the following theorem, which is without a doubt one of the most important results in Trigonometry.
Theorem 10.1. The Pythagorean Identity: For any angle θ, cos^2 (θ) + sin^2 (θ) = 1.
The moniker ‘Pythagorean’ brings to mind the Pythagorean Theorem, from which both the Distance Formula and the equation for a circle are ultimately derived.^5 The word ‘Identity’ reminds us that, regardless of the angle θ, the equation in Theorem 10.1 is always true. If one of cos(θ) or sin(θ) is known, Theorem 10.1 can be used to determine the other, up to a (±) sign. If, in addition, we know where the terminal side of θ lies when in standard position, then we can remove the ambiguity of the (±) and completely determine the missing value as the next example illustrates.
Example 10.2.2. Using the given information about θ, find the indicated value.
- If θ is a Quadrant II angle with sin(θ) = 35 , find cos(θ).
- If π < θ < 32 π with cos(θ) = −
√ 5 5 , find sin(θ).
- If sin(θ) = 1, find cos(θ).
Solution.
- When we substitute sin(θ) = 35 into The Pythagorean Identity, cos^2 (θ) + sin^2 (θ) = 1, we obtain cos^2 (θ) + 259 = 1. Solving, we find cos(θ) = ± 45. Since θ is a Quadrant II angle, its terminal side, when plotted in standard position, lies in Quadrant II. Since the x-coordinates are negative in Quadrant II, cos(θ) is too. Hence, cos(θ) = − 45.
- Substituting cos(θ) = −
√ 5 5 into cos
(^2) (θ) + sin (^2) (θ) = 1 gives sin(θ) = ± √ 2 5 =^ ±^
2 √ 5
- Since we are given that π < θ < 32 π , we know θ is a Quadrant III angle. Hence both its sine and cosine are negative and we conclude sin(θ) = − 2
√ 5
When we substitute sin(θ) = 1 into cos^2 (θ) + sin^2 (θ) = 1, we find cos(θ) = 0.
Another tool which helps immensely in determining cosines and sines of angles is the symmetry inherent in the Unit Circle. Suppose, for instance, we wish to know the cosine and sine of θ = 56 π. We plot θ in standard position below and, as usual, let P (x, y) denote the point on the terminal side of θ which lies on the Unit Circle. Note that the terminal side of θ lies π 6 radians short of one
half revolution. In Example 10.2.1, we determined that cos
( (^) π 6
√ 3 2 and sin^
( (^) π 6
= 12. This means (^4) This is unfortunate from a ‘function notation’ perspective. See Section 10.6. (^5) See Sections 1.1 and 7.2 for details.
10.2 The Unit Circle: Cosine and Sine 721
that the point on the terminal side of the angle π 6 , when plotted in standard position, is
3 2 ,^
1 2
From the figure below, it is clear that the point P (x, y) we seek can be obtained by reflecting that point about the y-axis. Hence, cos
( (^5) π 6
√ 3 2 and sin^
( (^5) π 6
x
y
1
1
P (x, y) θ = 56 π
π 6 x
y
1
1
“ √ 3 2 ,^ 1 2
” P
“ −
√ 3 2 ,^ 1 2
”
π 6 π 6
θ = 56 π
In the above scenario, the angle π 6 is called the reference angle for the angle 56 π. In general, for a non-quadrantal angle θ, the reference angle for θ (usually denoted α) is the acute angle made between the terminal side of θ and the x-axis. If θ is a Quadrant I or IV angle, α is the angle between the terminal side of θ and the positive x-axis; if θ is a Quadrant II or III angle, α is the angle between the terminal side of θ and the negative x-axis. If we let P denote the point (cos(θ), sin(θ)), then P lies on the Unit Circle. Since the Unit Circle possesses symmetry with respect to the x-axis, y-axis and origin, regardless of where the terminal side of θ lies, there is a point Q symmetric with P which determines θ’s reference angle, α as seen below.
x
y
1
1 P = Q
α
x
y
1
1 P Q
α α
Reference angle α for a Quadrant I angle Reference angle α for a Quadrant II angle
10.2 The Unit Circle: Cosine and Sine 723
sin(θ) < 0. The Reference Angle Theorem yields: cos (225◦) = − cos (45◦) = −
√ 2 2 and sin (225◦) = − sin (45◦) = −
√ 2
The terminal side of θ = 116 π , when plotted in standard position, lies in Quadrant IV, just shy of the positive x-axis. To find θ’s reference angle α, we subtract: α = 2π − θ = 2π − 116 π = π 6. Since θ is a Quadrant IV angle, cos(θ) > 0 and sin(θ) < 0, so the Reference Angle Theorem gives: cos
( (^11) π 6
= cos
( (^) π 6
√ 3 2 and sin^
( (^11) π 6
= − sin
( (^) π 6
x
y
1
1
θ = 225◦
45 ◦
Finding cos (225◦) and sin (225◦)
x
y
1
1
θ = 116 π
π 6
Finding cos
( (^11) π 6
and sin
( (^11) π 6
- To plot θ = − 54 π , we rotate clockwise an angle of 54 π from the positive x-axis. The terminal side of θ, therefore, lies in Quadrant II making an angle of α = 54 π − π = π 4 radians with respect to the negative x-axis. Since θ is a Quadrant II angle, the Reference Angle Theorem gives: cos
− 54 π
= − cos
( (^) π 4
√ 2 2 and sin^
− 54 π
= sin
( (^) π 4
√ 2
Since the angle θ = 73 π measures more than 2π = 63 π , we find the terminal side of θ by rotating one full revolution followed by an additional α = 73 π − 2 π = π 3 radians. Since θ and α are coterminal, cos
( (^7) π 3
= cos
( (^) π 3
= 12 and sin
( (^7) π 3
= sin
( (^) π 3
√ 3
x
y
1
1
θ = − 54 π
π 4
Finding cos
− 54 π
and sin
− 54 π
x
y
1
1
θ = 73 π^ π 3
Finding cos
( (^7) π 3
and sin
( (^7) π 3
724 Foundations of Trigonometry
The reader may have noticed that when expressed in radian measure, the reference angle for a non-quadrantal angle is easy to spot. Reduced fraction multiples of π with a denominator of 6 have π 6 as a reference angle, those with a denominator of 4 have π 4 as their reference angle, and those with a denominator of 3 have π 3 as their reference angle.^6 The Reference Angle Theorem in conjunction with the table of cosine and sine values on Page 722 can be used to generate the following figure, which the authors feel should be committed to memory.
x
y
(0, 1)
2 2 ,
√ 2 2
3 2 ,^
1 2
1 2 ,
√ 3 2
√ 2 2 ,
√ 2 2
√ 3 2 ,^
1 2
√ 3 2
2 2 ,^ −
√ 2 2
3 2 ,^ −^
1 2
1 2 ,^ −
√ 3 2
√ 2 2 ,^ −
√ 2 2
√ 3 2 ,^ −^
1 2
√ 3 2
0 , 2 π
π 2
π
3 π 2
π 4 π 6
π 3 3 π 4 5 π 6
2 π 3
5 π 4
7 π 6
4 π 3
7 π 4
11 π 6
5 π 3
Important Points on the Unit Circle (^6) For once, we have something convenient about using radian measure in contrast to the abstract theoretical nonsense about using them as a ‘natural’ way to match oriented angles with real numbers!
726 Foundations of Trigonometry
x
y
1
1 θ π
α
Visualizing θ = π + α
x
y
1
1 θ
α
θ has reference angle α
(b) Rewriting θ = 2π − α as θ = 2π + (−α), we can plot θ by visualizing one complete revolution counter-clockwise followed by a clockwise revolution, or ‘backing up,’ of α radians. We see that α is θ’s reference angle, and since θ is a Quadrant IV angle, the Reference Angle Theorem gives: cos(θ) = 135 and sin(θ) = − 1213.
x
y
1
1 θ
2 π −α
Visualizing θ = 2π − α
x
y
1
1 θ
α
θ has reference angle α
(c) Taking a cue from the previous problem, we rewrite θ = 3π − α as θ = 3π + (−α). The angle 3π represents one and a half revolutions counter-clockwise, so that when we ‘back up’ α radians, we end up in Quadrant II. Using the Reference Angle Theorem, we get cos(θ) = − 135 and sin(θ) = 1213.
10.2 The Unit Circle: Cosine and Sine 727
x
y
1
1
−α 3 π
Visualizing 3π − α
x
y
1
1 θ α
θ has reference angle α
(d) To plot θ = π 2 + α, we first rotate π 2 radians and follow up with α radians. The reference angle here is not α, so The Reference Angle Theorem is not immediately applicable. (It’s important that you see why this is the case. Take a moment to think about this before reading on.) Let Q(x, y) be the point on the terminal side of θ which lies on the Unit Circle so that x = cos(θ) and y = sin(θ). Once we graph α in standard position, we use the fact that equal angles subtend equal chords to show that the dotted lines in the figure below are equal. Hence, x = cos(θ) = − 1213. Similarly, we find y = sin(θ) = 135.
x
y
1
1 θ π 2 α
Visualizing θ = π 2 + α
x
y
1
1 P
13 ,^
12 13
Q (x, y) (^) α
α
Using symmetry to determine Q(x, y)
10.2 The Unit Circle: Cosine and Sine 729
x
y
1
− (^12)
1
π 6
x
y
1
− (^12)
1
π 6
In Quadrant III, one solution is 76 π , so we capture all Quadrant III solutions by adding integer multiples of 2π: θ = 76 π + 2πk. In Quadrant IV, one solution is 116 π so all the solutions here are of the form θ = 116 π + 2πk for integers k.
- The angles with cos(θ) = 0 are quadrantal angles whose terminal sides, when plotted in standard position, lie along the y-axis.
x
y
1
1
π 2
x
y
1
1
π 2
π 2 π
While, technically speaking, π 2 isn’t a reference angle we can nonetheless use it to find our answers. If we follow the procedure set forth in the previous examples, we find θ = π 2 + 2πk and θ = 32 π + 2πk for integers, k. While this solution is correct, it can be shortened to θ = π 2 + πk for integers k. (Can you see why this works from the diagram?)
One of the key items to take from Example 10.2.5 is that, in general, solutions to trigonometric equations consist of infinitely many answers. To get a feel for these answers, the reader is encouraged to follow our mantra from Chapter 9 - that is, ‘When in doubt, write it out!’ This is especially important when checking answers to the exercises. For example, another Quadrant IV solution to sin(θ) = − 12 is θ = − π 6. Hence, the family of Quadrant IV answers to number 2 above could just have easily been written θ = − π 6 + 2πk for integers k. While on the surface, this family may look
730 Foundations of Trigonometry
different than the stated solution of θ = 116 π + 2πk for integers k, we leave it to the reader to show they represent the same list of angles.
10.2.1 Beyond the Unit Circle
We began the section with a quest to describe the position of a particle experiencing circular motion. In defining the cosine and sine functions, we assigned to each angle a position on the Unit Circle. In this subsection, we broaden our scope to include circles of radius r centered at the origin. Consider for the moment the acute angle θ drawn below in standard position. Let Q(x, y) be the point on the terminal side of θ which lies on the circle x^2 + y^2 = r^2 , and let P (x′, y′) be the point on the terminal side of θ which lies on the Unit Circle. Now consider dropping perpendiculars from P and Q to create two right triangles, ∆OP A and ∆OQB. These triangles are similar,^10 thus it follows that (^) xx′ = r 1 = r, so x = rx′^ and, similarly, we find y = ry′. Since, by definition, x′^ = cos(θ) and y′^ = sin(θ), we get the coordinates of Q to be x = r cos(θ) and y = r sin(θ). By reflecting these points through the x-axis, y-axis and origin, we obtain the result for all non-quadrantal angles θ, and we leave it to the reader to verify these formulas hold for the quadrantal angles.
x
y
1
1
r
r Q (x, y)
P (x′, y′) θ
θ
x
y
1
O A(x′, 0) B(x, 0)
P (x′, y′)
Q(x, y) = (r cos(θ), r sin(θ))
Not only can we describe the coordinates of Q in terms of cos(θ) and sin(θ) but since the radius of the circle is r =
x^2 + y^2 , we can also express cos(θ) and sin(θ) in terms of the coordinates of Q. These results are summarized in the following theorem.
Theorem 10.3. If Q(x, y) is the point on the terminal side of an angle θ, plotted in standard position, which lies on the circle x^2 + y^2 = r^2 then x = r cos(θ) and y = r sin(θ). Moreover,
cos(θ) = x r
x √ x^2 + y^2
and sin(θ) = y r
y √ x^2 + y^2
(^10) Do you remember why?
732 Foundations of Trigonometry
Theorem 10.3 gives us what we need to describe the position of an object traveling in a circular path of radius r with constant angular velocity ω. Suppose that at time t, the object has swept out an angle measuring θ radians. If we assume that the object is at the point (r, 0) when t = 0, the angle θ is in standard position. By definition, ω = θt which we rewrite as θ = ωt. According to Theorem 10.3, the location of the object Q(x, y) on the circle is found using the equations x = r cos(θ) = r cos(ωt) and y = r sin(θ) = r sin(ωt). Hence, at time t, the object is at the point (r cos(ωt), r sin(ωt)). We have just argued the following.
Equation 10.3. Suppose an object is traveling in a circular path of radius r centered at the origin with constant angular velocity ω. If t = 0 corresponds to the point (r, 0), then the x and y coordinates of the object are functions of t and are given by x = r cos(ωt) and y = r sin(ωt). Here, ω > 0 indicates a counter-clockwise direction and ω < 0 indicates a clockwise direction.
x
y
1
1
r
r Q (x, y) = (r cos(ωt), r sin(ωt))
θ = ωt
Equations for Circular Motion
Example 10.2.7. Suppose we are in the situation of Example 10.1.5. Find the equations of motion of Lakeland Community College as the earth rotates.
Solution. From Example 10.1.5, we take r = 2960 miles and and ω = (^) 12 hoursπ. Hence, the equations of motion are x = r cos(ωt) = 2960 cos
( (^) π 12 t
and y = r sin(ωt) = 2960 sin
( (^) π 12 t
, where x and y are measured in miles and t is measured in hours.
In addition to circular motion, Theorem 10.3 is also the key to developing what is usually called ‘right triangle’ trigonometry.^11 As we shall see in the sections to come, many applications in trigonometry involve finding the measures of the angles in, and lengths of the sides of, right triangles. Indeed, we made good use of some properties of right triangles to find the exact values of the cosine and sine of many of the angles in Example 10.2.1, so the following development shouldn’t be that much of a surprise. Consider the generic right triangle below with corresponding acute angle θ. The side with length a is called the side of the triangle adjacent to θ; the side with length b is called the side of the triangle opposite θ; and the remaining side of length c (the side opposite the
(^11) You may have been exposed to this in High School.
10.2 The Unit Circle: Cosine and Sine 733
right angle) is called the hypotenuse. We now imagine drawing this triangle in Quadrant I so that the angle θ is in standard position with the adjacent side to θ lying along the positive x-axis.
θ a
b
c x
y
c
c
P (a, b) θ
According to the Pythagorean Theorem, a^2 + b^2 = c^2 , so that the point P (a, b) lies on a circle of radius c. Theorem 10.3 tells us that cos(θ) = ac and sin(θ) = bc , so we have determined the cosine and sine of θ in terms of the lengths of the sides of the right triangle. Thus we have the following theorem.
Theorem 10.4. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hypotenuse is c, then cos(θ) =
a c and sin(θ) =
b c
Example 10.2.8. Find the measure of the missing angle and the lengths of the missing sides of:
Solution. The first and easiest task is to find the measure of the missing angle. Since the sum of angles of a triangle is 180◦, we know that the missing angle has measure 180◦^ − 30 ◦^ − 90 ◦^ = 60◦. We now proceed to find the lengths of the remaining two sides of the triangle. Let c denote the length of the hypotenuse of the triangle. By Theorem 10.4, we have cos (30◦) = (^7) c , or c = (^) cos(30^7 ◦).
Since cos (30◦) =
√ 3 2 , we have, after the usual fraction gymnastics,^ c^ =^
14 √ 3 3.^ At this point, we have two ways to proceed to find the length of the side opposite the 30◦^ angle, which we’ll denote b. We know the length of the adjacent side is 7 and the length of the hypotenuse is 14
√ 3 3 , so we
10.2 The Unit Circle: Cosine and Sine 735
Theorem 10.5. Domain and Range of the Cosine and Sine Functions:
- The function f (t) = cos(t) • The function g(t) = sin(t)
- has domain (−∞, ∞) – has domain (−∞, ∞)
- has range [− 1 , 1] – has range [− 1 , 1]
Suppose, as in the Exercises, we are asked to solve an equation such as sin(t) = − 12. As we have already mentioned, the distinction between t as a real number and as an angle θ = t radians is often blurred. Indeed, we solve sin(t) = − 12 in the exact same manner^12 as we did in Example 10.2. number 2. Our solution is only cosmetically different in that the variable used is t rather than θ: t = 76 π + 2πk or t = 116 π + 2πk for integers, k. We will study the cosine and sine functions in greater detail in Section 10.5. Until then, keep in mind that any properties of cosine and sine developed in the following sections which regard them as functions of angles in radian measure apply equally well if the inputs are regarded as real numbers.
(^12) Well, to be pedantic, we would be technically using ‘reference numbers’ or ‘reference arcs’ instead of ‘reference angles’ – but the idea is the same.
736 Foundations of Trigonometry
10.2.2 Exercises
In Exercises 1 - 20, find the exact value of the cosine and sine of the given angle.
- θ = 0 2. θ = π 4 3. θ = π 3 4. θ = π 2
- θ =
2 π 3
- θ =
3 π 4
- θ = π 8. θ =
7 π 6
- θ = 5 π 4 10. θ = 4 π 3 11. θ = 3 π 2 12. θ = 5 π 3
- θ =
7 π 4
- θ =
23 π 6
- θ = −
13 π 2
- θ = −
43 π 6
- θ = − 3 π 4 18. θ = − π 6 19. θ = 10 π 3 20. θ = 117π
In Exercises 21 - 30, use the results developed throughout the section to find the requested value.
- If sin(θ) = −
with θ in Quadrant IV, what is cos(θ)?
- If cos(θ) =
with θ in Quadrant I, what is sin(θ)?
- If sin(θ) =
with θ in Quadrant II, what is cos(θ)?
- If cos(θ) = −
with θ in Quadrant III, what is sin(θ)?
- If sin(θ) = −
with θ in Quadrant III, what is cos(θ)?
- If cos(θ) =
with θ in Quadrant IV, what is sin(θ)?
- If sin(θ) =
and π 2
< θ < π, what is cos(θ)?
- If cos(θ) =
and 2π < θ < 5 π 2
, what is sin(θ)?
- If sin(θ) = − 0 .42 and π < θ <
3 π 2 , what is cos(θ)?
- If cos(θ) = − 0 .98 and
π 2 < θ < π, what is sin(θ)?
