7.3 Unit Circle
Figure 1 The Singapore Flyer was the world’s tallest Ferris wheel until being overtaken by the High Roller in Las Vegas and the Ain Dubai in Dubai. (credit: ʺVibin JKʺ/Flickr)
Looking for a thrill? Then consider a ride on the Ain Dubai, the world’s tallest Ferris wheel. Located in Dubai, the most populous city and the financial and tourism hub of the United Arab Emirates, the wheel soars to 820 feet, about 1.5 tenths of a mile. Described as an observation wheel, riders enjoy spectacular views of the Burj Khalifa (the world’s tallest building) and the Palm Jumeirah (a human-made archipelago home to over 10,000 people and 20 resorts) as they travel from the ground to the peak and down again in a repeating pattern. In this section, we will examine this type of revolving motion around a circle. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. Then we can discuss circular motion in terms of the coordinate pairs.
7.3.1 Finding Trigonometric Functions Using the Unit Circle
We have already defined the trigonometric functions in terms of right triangles. In this section, we will redefine them in terms of the unit circle. Recall that a unit circle is a circle centered at the origin with radius 1, as shown in Figure 2. The angle (in radians) that \(t\) intercepts forms an arc of length \(s.\) Using the formula \(s = rt,\) and knowing that \(r = 1,\) we see that for a unit circle, \(s = t.\)
The \(x\)- and \(y\)-axes divide the coordinate plane into four quarters called quadrants. We label these quadrants to mimic the direction a positive angle would sweep. The four quadrants are labeled I, II, III, and IV.
For any angle \(t,\) we can label the intersection of the terminal side and the unit circle as by its coordinates, \(\left( {x,y} \right).\) The coordinates \(x\) and \(y\) will be the outputs of the trigonometric functions \(f(t) = \cos\; t\) and \(f(t) = \sin\; t,\) respectively. This means \(\; x = \text{cos~}t\;\) and \(\; y = \text{sin~}t.\)
Figure 2 Unit circle where the central angle is \(t\) radians
Defining Sine and Cosine Functions from the Unit Circle
The sine function relates a real number \(t\) to the \(y\)-coordinate of the point where the corresponding angle intercepts the unit circle. More precisely, the sine of an angle \(t\) equals the \(y\)-value of the endpoint on the unit circle of an arc of length \(t.\) In Figure 2, the sine is equal to \(y.\) Like all functions, the sine function has an input and an output. Its input is the measure of the angle; its output is the \(y\)-coordinate of the corresponding point on the unit circle.
The cosine function of an angle \(t\) equals the \(x\)-value of the endpoint on the unit circle of an arc of length \(t.\) In Figure 3, the cosine is equal to \(x.\)
Figure 3
Because it is understood that sine and cosine are functions, we do not always need to write them with parentheses: \(\sin\; t\) is the same as \(\sin(t)\) and \(\cos t\) is the same as \(\cos(t).\) Likewise, \(\cos^{2}t\) is a commonly used shorthand notation for \({(\cos(t))}^{2}.\) Be aware that many calculators and computers do not recognize the shorthand notation. When in doubt, use the extra parentheses when entering calculations into a calculator or computer.
Point \(P\) is a point on the unit circle corresponding to an angle of \(t,\) as shown in Figure 4. Find \(\cos(t)\) and \(\text{sin}(t).\)
Figure 4
Solution (click to reveal)
We know that \(\cos\; t\) is the \(x\)-coordinate of the corresponding point on the unit circle and \(\sin\; t\) is the \(y\)-coordinate of the corresponding point on the unit circle. So:
\[\begin{array}{cll} x & {= \cos\; t} & {= \frac{1}{2}} \\ y & {= \sin\; t} & {= \frac{\sqrt{3}}{2}} \end{array}\]
A certain angle \(t\) corresponds to a point on the unit circle at \(\left( {- \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}} \right)\) as shown in Figure 5. Find \(\cos\; t\) and \(\sin\; t.\)
Figure 5
Solution (click to reveal)
\(\cos(t) = - \frac{\sqrt{2}}{2},\sin(t) = \frac{\sqrt{2}}{2}\)
Finding Sines and Cosines of Angles on an Axis
For quadrantral angles, the corresponding point on the unit circle falls on the \(x\)- or \(y\)-axis. In that case, we can easily calculate cosine and sine from the values of \(x\) and \(y.\)
Find \(\text{cos}(90{^\circ})\) and \(\text{sin}(90{^\circ}).\)
Solution (click to reveal)
Moving \(90{^\circ}\) counterclockwise around the unit circle from the positive \(x\)-axis brings us to the top of the circle, where the \((x,y)\) coordinates are \((0,1),\) as shown in Figure 6.
Figure 6
We can then use our definitions of cosine and sine.
\[\begin{array}{rlll} x & {= \text{cos~}t} & {= \cos(90{^\circ})} & {= 0} \\ y & {= \text{sin~}t} & {= \sin(90{^\circ})} & {= 1} \end{array}\]
The cosine of \(90{^\circ}\) is 0; the sine of \(90{^\circ}\) is 1.
The Pythagorean Identity
Now that we can define sine and cosine, we will learn how they relate to each other and the unit circle. Recall that the equation for the unit circle is \(x^{2} + y^{2} = 1.\) Because \(x = \cos\; t\) and \(y = \sin\; t,\) we can substitute for \(x\) and \(y\) to get \(\cos^{2}t + \sin^{2}t = 1.\) This equation, \(\cos^{2}t + \sin^{2}t = 1,\) is known as the Pythagorean Identity. See Figure 7.
Figure 7
We can use the Pythagorean Identity to find the cosine of an angle if we know the sine, or vice versa. However, because the equation yields two solutions, we need additional knowledge of the angle to choose the solution with the correct sign. If we know the quadrant where the angle is, we can easily choose the correct solution.
If \(\sin(t) = \frac{3}{7}\) and \(t\) is in the second quadrant, find \(\cos(t).\)
Solution (click to reveal)
If we drop a vertical line from the point on the unit circle corresponding to \(t,\) we create a right triangle, from which we can see that the Pythagorean Identity is simply one case of the Pythagorean Theorem. See Figure 8.
Figure 8
Substituting the known value for sine into the Pythagorean Identity,
\[\begin{array}{rcl} {\text{cos}^{2}(t) + \sin^{2}(t)} & = & 1 \\ {\text{cos}^{2}(t) + \frac{9}{49}} & = & 1 \\ {\text{cos}^{2}(t)} & = & \frac{40}{49} \\ {\text{cos}(t)} & = & {\pm \sqrt{\frac{40}{49}} = \pm \frac{\sqrt{40}}{7} = \pm \frac{2\sqrt{10}}{7}} \end{array}\]
Because the angle is in the second quadrant, we know the \(x\)-value is a negative real number, so the cosine is also negative.
\[\text{cos}(t) = - \frac{2\sqrt{10}}{7}\]
7.3.2 Finding Sines and Cosines of Special Angles
We have already learned some properties of the special angles, such as the conversion from radians to degrees, and we found their sines and cosines using right triangles. We can also calculate sines and cosines of the special angles using the Pythagorean Identity.
Finding Sines and Cosines of \(45{^\circ}\) Angles
First, we will look at angles of \(45{^\circ}\) or \(\frac{\pi}{4},\) as shown in Figure 9. A \(45{^\circ}–45{^\circ}–90{^\circ}\) triangle is an isosceles triangle, so the \(x\)- and \(y\)-coordinates of the corresponding point on the circle are the same. Because the \(x\)- and \(y\)-values are the same, the sine and cosine values will also be equal.
Figure 9
At \(t = \frac{\pi}{4},\) which is 45 degrees, the radius of the unit circle bisects the first quadrantal angle. This means the radius lies along the line \(y = x.\) A unit circle has a radius equal to 1 so the right triangle formed below the line \(y = x\) has sides \(x\) and \(y\ (y = x),\) and radius = 1. See Figure 10.
Figure 10
From the Pythagorean Theorem we get
\[x^{2} + y^{2} = 1\]
We can then substitute \(y = x.\)
\[x^{2} + x^{2} = 1\]
Next we combine like terms.
\[2x^{2} = 1\]
And solving for \(x,\) we get
\[\begin{array}{rcl} x^{2} & = & \frac{1}{2} \\ x & = & {\pm \frac{1}{\sqrt{2}}} \end{array}\]
In quadrant I, \(x = \frac{1}{\sqrt{2}}.\)
At \(t = \frac{\pi}{4}\) or 45 degrees,
\[\begin{array}{rcl} \left( {x,y} \right) & = & {\left( {x,x} \right) = \left( {\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}} \right)} \\ x & = & {\frac{1}{\sqrt{2}},y = \frac{1}{\sqrt{2}}} \\ {\text{cos~}t} & = & {\frac{1}{\sqrt{2}},\text{sin~}t = \frac{1}{\sqrt{2}}} \end{array}\]
If we then rationalize the denominators, we get
\[\begin{matrix} {\text{cos~}t} & = & {\frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}} \\ & = & \frac{\sqrt{2}}{2} \\ {\text{sin~}t} & = & {\frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}} \\ & = & \frac{\sqrt{2}}{2} \end{matrix}\]
Therefore, the \((x,y)\) coordinates of a point on a circle of radius \(1\) at an angle of \(45{^\circ}\) are \(\left( {\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}} \right).\)
Finding Sines and Cosines of \(30{^\circ}\) and \(60{^\circ}\) Angles
Next, we will find the cosine and sine at an angle of \(30{^\circ},\) or \(\frac{\pi}{6}.\) First, we will draw a triangle inside a circle with one side at an angle of \(30{^\circ},\) and another at an angle of \(-30{^\circ},\) as shown in Figure 11. If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be \(60{^\circ},\) as shown in Figure 12.
Figure 11
Figure 12
Because all the angles are equal, the sides are also equal. The vertical line has length \(2y,\) and since the sides are all equal, we can also conclude that \(r = 2y\) or \(y = \frac{1}{2}r.\) Since \(\sin\; t = y,\)
\[\sin\left( \frac{\pi}{6} \right) = \frac{1}{2}r\]
And since \(r = 1\) in our unit circle,
\[\begin{matrix} {\sin\left( \frac{\pi}{6} \right)} & = & {\frac{1}{2}(1)} \\ & = & \frac{1}{2} \end{matrix}\]
Using the Pythagorean Identity, we can find the cosine value.
\[\begin{array}{rclc} {\cos^{2}\left( \frac{\pi}{6} \right) + \sin^{2}\left( \frac{\pi}{6} \right)} & = & 1 & \\ {\cos^{2}\left( \frac{\pi}{6} \right) + \left( \frac{1}{2} \right)^{2}} & = & 1 & \\ {\cos^{2}\left( \frac{\pi}{6} \right)} & = & \frac{3}{4} & {\quad\text{Use~the~square~root~property}.} \\ {\cos\left( \frac{\pi}{6} \right)} & = & {\frac{\pm \sqrt{3}}{\pm \sqrt{4}} = \frac{\sqrt{3}}{2}} & {\quad\text{Since~}y\text{~is~positive,~choose~the~positive~root}.} \end{array}\]
The \((x,y)\) coordinates for the point on a circle of radius \(1\) at an angle of \(30{^\circ}\) are \(\left( {\frac{\sqrt{3}}{2},\frac{1}{2}} \right).\) At \(t = \frac{\pi}{3}\text{~(60°}\text{),}\) the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, \(BAD,\) as shown in Figure 13. Angle \(A\) has measure \(60{^\circ}.\) At point \(B,\) we draw an angle \(ABC\) with measure of \(60{^\circ}.\) We know the angles in a triangle sum to \(180{^\circ},\) so the measure of angle \(C\) is also \(60{^\circ}.\) Now we have an equilateral triangle. Because each side of the equilateral triangle \(ABC\) is the same length, and we know one side is the radius of the unit circle, all sides must be of length 1.
Figure 13
The measure of angle \(ABD\) is 30°. Angle \(ABC\) is double angle \(ABD,\) so its measure is 60°. \(BD\) is the perpendicular bisector of \(AC,\) so it cuts \(AC\) in half. This means that \(AD\) is \(\frac{1}{2}\) the radius, or \(\frac{1}{2}.\) Notice that \(AD\) is the \(x\)-coordinate of point \(B,\) which is at the intersection of the 60° angle and the unit circle. This gives us a triangle \(BAD\) with hypotenuse of 1 and side \(x\) of length \(\frac{1}{2}.\)
From the Pythagorean Theorem, we get
\[x^{2} + y^{2} = 1\]
Substituting \(x = \frac{1}{2},\) we get
\[\left( \frac{1}{2} \right)^{2} + y^{2} = 1\]
Solving for \(y,\) we get
\[\begin{array}{rcl} {\frac{1}{4} + y^{2}} & = & 1 \\ y^{2} & = & {1 - \frac{1}{4}} \\ y^{2} & = & \frac{3}{4} \\ y & = & {\pm \frac{\sqrt{3}}{2}} \end{array}\]
Since \(t = \frac{\pi}{3}\) has the terminal side in quadrant I where the \(y\)-coordinate is positive, we choose \(y = \frac{\sqrt{3}}{2},\) the positive value.
At \(t = \frac{\pi}{3}\) (60°), the \((x,y)\) coordinates for the point on a circle of radius \(1\) at an angle of \(60{^\circ}\) are \(\left( {\frac{1}{2},\frac{\sqrt{3}}{2}} \right),\) so we can find the sine and cosine.
\[\begin{array}{rcl} {(x,y)} & = & \left( {\frac{1}{2},\frac{\sqrt{3}}{2}} \right) \\ x & = & {\frac{1}{2},y = \frac{\sqrt{3}}{2}} \\ {\text{cos~}t} & = & {\frac{1}{2},\text{sin~}t = \frac{\sqrt{3}}{2}} \end{array}\]
We have now found the cosine and sine values for all of the most commonly encountered angles in the first quadrant of the unit circle. Table 1 summarizes these values.
| Angle | \(0\) | \(\frac{\pi}{6},\) or \(30{^\circ}\) | \(\frac{\pi}{4},\) or \(45{^\circ}\) | \(\frac{\pi}{3},\) or \(60{^\circ}\) | \(\frac{\pi}{2},\) or \(90{^\circ}\) |
| Cosine | 1 | \(\frac{\sqrt{3}}{2}\) | \(\frac{\sqrt{2}}{2}\) | \(\frac{1}{2}\) | 0 |
| Sine | 0 | \(\frac{1}{2}\) | \(\frac{\sqrt{2}}{2}\) | \(\frac{\sqrt{3}}{2}\) | 1 |
Table 1
Figure 14 shows the common angles in the first quadrant of the unit circle.
Figure 14
Using a Calculator to Find Sine and Cosine
To find the cosine and sine of angles other than the special angles, we turn to a computer or calculator. Be aware: Most calculators can be set into “degree” or “radian” mode, which tells the calculator the units for the input value. When we evaluate \(\cos(30)\) on our calculator, it will evaluate it as the cosine of 30 degrees if the calculator is in degree mode, or the cosine of 30 radians if the calculator is in radian mode.
7.3.3 Identifying the Domain and Range of Sine and Cosine Functions
Now that we can find the sine and cosine of an angle, we need to discuss their domains and ranges. What are the domains of the sine and cosine functions? That is, what are the smallest and largest numbers that can be inputs of the functions? Because angles smaller than \(0\) and angles larger than \(2\pi\) can still be graphed on the unit circle and have real values of \(x,y,\) and \(r,\) there is no lower or upper limit to the angles that can be inputs to the sine and cosine functions. The input to the sine and cosine functions is the rotation from the positive \(x\)-axis, and that may be any real number.
What are the ranges of the sine and cosine functions? What are the least and greatest possible values for their output? We can see the answers by examining the unit circle, as shown in Figure 15. The bounds of the \(x\)-coordinate are \(\lbrack-1,1\rbrack.\) The bounds of the \(y\)-coordinate are also \(\lbrack-1,1\rbrack.\) Therefore, the range of both the sine and cosine functions is \(\lbrack-1,1\rbrack.\)
Figure 15
7.3.4 Finding Reference Angles
We have discussed finding the sine and cosine for angles in the first quadrant, but what if our angle is in another quadrant? For any given angle in the first quadrant, there is an angle in the second quadrant with the same sine value. Because the sine value is the \(y\)-coordinate on the unit circle, the other angle with the same sine will share the same \(y\)-value, but have the opposite \(x\)-value. Therefore, its cosine value will be the opposite of the first angle’s cosine value.
Likewise, there will be an angle in the fourth quadrant with the same cosine as the original angle. The angle with the same cosine will share the same \(x\)-value but will have the opposite \(y\)-value. Therefore, its sine value will be the opposite of the original angle’s sine value.
As shown in Figure 16, angle \(\alpha\) has the same sine value as angle \(t;\) the cosine values are opposites. Angle \(\beta\) has the same cosine value as angle \(t;\) the sine values are opposites.
\[\begin{array}{lcl} {\sin(t) = \sin(\alpha)} & {\quad\text{and}\quad} & {\cos(t) = - \cos(\alpha)} \\ {\sin(t) = - \sin(\beta)} & {\quad\text{and}\quad} & {\cos(t) = \cos(\beta)} \end{array}\]
Figure 16
Recall that an angle’s reference angle is the acute angle, \(t,\) formed by the terminal side of the angle \(t\) and the horizontal axis. A reference angle is always an angle between \(0\) and \(90{^\circ},\) or \(0\) and \(\frac{\pi}{2}\) radians. As we can see from Figure 17, for any angle in quadrants II, III, or IV, there is a reference angle in quadrant I.
Figure 17
Find the reference angle of \(225{^\circ}\) as shown in Figure 18.
Figure 18
Solution (click to reveal)
Because \(225{^\circ}\) is in the third quadrant, the reference angle is
\[\left| \left( {180{^\circ} - 225{^\circ}} \right) \right| = \left| {- 45{^\circ}} \right| = 45{^\circ}\]
7.3.5 Using Reference Angles
Now let’s take a moment to reconsider the Ferris wheel introduced at the beginning of this section. Suppose a rider snaps a photograph while stopped twenty feet above ground level. The rider then rotates three-quarters of the way around the circle. What is the rider’s new elevation? To answer questions such as this one, we need to evaluate the sine or cosine functions at angles that are greater than 90 degrees or at a negative angle. Reference angles make it possible to evaluate trigonometric functions for angles outside the first quadrant. They can also be used to find \(\left( {x,y} \right)\) coordinates for those angles. We will use the reference angle of the angle of rotation combined with the quadrant in which the terminal side of the angle lies.
Using Reference Angles to Evaluate Trigonometric Functions
We can find the cosine and sine of any angle in any quadrant if we know the cosine or sine of its reference angle. The absolute values of the cosine and sine of an angle are the same as those of the reference angle. The sign depends on the quadrant of the original angle. The cosine will be positive or negative depending on the sign of the \(x\)-values in that quadrant. The sine will be positive or negative depending on the sign of the \(y\)-values in that quadrant.
Using Reference Angles to Find Coordinates
Now that we have learned how to find the cosine and sine values for special angles in the first quadrant, we can use symmetry and reference angles to fill in cosine and sine values for the rest of the special angles on the unit circle. They are shown in Figure 19. Take time to learn the \((x,y)\) coordinates of all of the major angles in the first quadrant.
Figure 19 Special angles and coordinates of corresponding points on the unit circle
In addition to learning the values for special angles, we can use reference angles to find \(\left( {x,y} \right)\) coordinates of any point on the unit circle, using what we know of reference angles along with the identities
\[\begin{array}{l} {x = \text{cos~}t} \\ {y = \text{sin~}t} \end{array}\]
First we find the reference angle corresponding to the given angle. Then we take the sine and cosine values of the reference angle, and give them the signs corresponding to the \(y\)- and \(x\)-values of the quadrant.
Section Exercises
Verbal
1. Describe the unit circle.
Solution (click to reveal)
The unit circle is a circle of radius 1 centered at the origin.
2. What do the \(x\)- and \(y\)-coordinates of the points on the unit circle represent?
3. Discuss the difference between a coterminal angle and a reference angle.
Solution (click to reveal)
Coterminal angles are angles that share the same terminal side. A reference angle is the size of the smallest acute angle, \(t,\) formed by the terminal side of the angle \(t\) and the horizontal axis.
4. Explain how the cosine of an angle in the second quadrant differs from the cosine of its reference angle in the unit circle.
5. Explain how the sine of an angle in the second quadrant differs from the sine of its reference angle in the unit circle.
Solution (click to reveal)
The sine values are equal.
Algebraic
For the following exercises, use the given sign of the sine and cosine functions to find the quadrant in which the terminal point determined by \(t\) lies.
6. \(\text{sin}(t) < 0\) and \(\text{cos}(t) < 0\)
7. \(\text{sin}(t) > 0\) and \(\cos(t) > 0\)
Solution (click to reveal)
I
8. \(\text{sin}(t) > 0\) and \(\cos(t) < 0\)
9. \(\text{sin}(t) < 0\) and \(\cos(t) > 0\)
Solution (click to reveal)
IV
For the following exercises, find the exact value of each trigonometric function.
10. \(\sin\;\frac{\pi}{2}\)
11. \(\sin\;\frac{\pi}{3}\)
Solution (click to reveal)
\(\frac{\sqrt{3}}{2}\)
12. \(\cos\;\frac{\pi}{2}\)
13. \(\cos\;\frac{\pi}{3}\)
Solution (click to reveal)
\(\frac{1}{2}\)
14. \(\sin\;\frac{\pi}{4}\)
15. \(\cos\;\frac{\pi}{4}\)
Solution (click to reveal)
\(\frac{\sqrt{2}}{2}\)
16. \(\sin\;\frac{\pi}{6}\)
17. \(\sin\;\pi\)
Solution (click to reveal)
0
18. \(\sin\;\frac{3\pi}{2}\)
19. \(\cos\;\pi\)
Solution (click to reveal)
-1
20. \(\cos\; 0\)
21. \(\cos\;\frac{\pi}{6}\)
Solution (click to reveal)
\(\frac{\sqrt{3}}{2}\)
22. \(\sin\; 0\)
Numeric
For the following exercises, state the reference angle for the given angle.
23. \(240{^\circ}\)
Solution (click to reveal)
\(60{^\circ}\)
24. \(-170{^\circ}\)
25. \(100{^\circ}\)
Solution (click to reveal)
\(80{^\circ}\)
26. \(-315{^\circ}\)
27. \(135{^\circ}\)
Solution (click to reveal)
\(45{^\circ}\)
28. \(\frac{5\pi}{4}\)
29. \(\frac{2\pi}{3}\)
Solution (click to reveal)
\(\frac{\pi}{3}\)
30. \(\frac{5\pi}{6}\)
31. \(\frac{- 11\pi}{3}\)
Solution (click to reveal)
\(\frac{\pi}{3}\)
32. \(\frac{-7\pi}{4}\)
33. \(\frac{- \pi}{8}\)
Solution (click to reveal)
\(\frac{\pi}{8}\)
For the following exercises, find the reference angle, the quadrant of the terminal side, and the sine and cosine of each angle. If the angle is not one of the angles on the unit circle, use a calculator and round to three decimal places.
34. \(225{^\circ}\)
35. \(300{^\circ}\)
Solution (click to reveal)
\(60{^\circ},\) Quadrant IV, \(\text{sin}(300{^\circ}) = - \frac{\sqrt{3}}{2}\),\(\cos(300{^\circ}) = \frac{1}{2}\)
36. \(320{^\circ}\)
37. \(135{^\circ}\)
Solution (click to reveal)
\(45{^\circ},\) Quadrant II, \(\text{sin}(135{^\circ}) = \frac{\sqrt{2}}{2}\),\(\cos(135{^\circ}) = - \frac{\sqrt{2}}{2}\)
38. \(210{^\circ}\)
39. \(120{^\circ}\)
Solution (click to reveal)
\(60{^\circ},\) Quadrant II, \(\text{sin}(120{^\circ}) = \frac{\sqrt{3}}{2}\),\(\cos(120{^\circ}) = - \frac{1}{2}\)
40. \(250{^\circ}\)
41. \(150{^\circ}\)
Solution (click to reveal)
\(30{^\circ},\) Quadrant II, \(\text{sin}(150{^\circ}) = \frac{1}{2}\),\(\cos(150{^\circ}) = - \frac{\sqrt{3}}{2}\)
42. \(\frac{5\pi}{4}\)
43. \(\frac{7\pi}{6}\)
Solution (click to reveal)
\(\frac{\pi}{6},\) Quadrant III, \(\text{sin}\left( \frac{7\pi}{6} \right) = - \frac{1}{2}\),\(\cos\left( \frac{7\pi}{6} \right) = - \frac{\sqrt{3}}{2}\)
44. \(\frac{5\pi}{3}\)
45. \(\frac{3\pi}{4}\)
Solution (click to reveal)
\(\frac{\pi}{4},\) Quadrant II, \(\text{sin}\left( \frac{3\pi}{4} \right) = \frac{\sqrt{2}}{2}\),\(\cos\left( \frac{3\pi}{4} \right) = - \frac{\sqrt[{}]{2}}{2}\)
46. \(\frac{4\pi}{3}\)
47. \(\frac{2\pi}{3}\)
Solution (click to reveal)
\(\frac{\pi}{3},\) Quadrant II, \(\text{sin}\left( \frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2}\),\(\cos\left( \frac{2\pi}{3} \right) = - \frac{1}{2}\)
48. \(\frac{5\pi}{6}\)
49. \(\frac{7\pi}{4}\)
Solution (click to reveal)
\(\frac{\pi}{4},\) Quadrant IV, \(\text{sin}\left( \frac{7\pi}{4} \right) = - \frac{\sqrt{2}}{2},\text{cos}\left( \frac{7\pi}{4} \right) = \frac{\sqrt{2}}{2}\)
For the following exercises, find the requested value.
50. If \(\text{cos}(t) = \frac{1}{7}\) and \(t\) is in the fourth quadrant, find \(\text{sin}(t).\)
51. If \(\text{cos}(t) = \frac{2}{9}\) and \(t\) is in the first quadrant, find \(\text{sin}(t).\)
Solution (click to reveal)
\(\frac{\sqrt{77}}{9}\)
52. If \(\text{sin}(t) = \frac{3}{8}\) and \(t\) is in the second quadrant, find \(\text{cos}(t).\)
53. If \(\text{sin}(t) = - \frac{1}{4}\) and \(t\) is in the third quadrant, find \(\text{cos}(t).\)
Solution (click to reveal)
\(- \frac{\sqrt{15}}{4}\)
54. Find the coordinates of the point on a circle with radius 15 corresponding to an angle of \(220{^\circ}.\)
55. Find the coordinates of the point on a circle with radius 20 corresponding to an angle of \(120{^\circ}.\)
Solution (click to reveal)
\(\left( {-10,\; 10\sqrt{3}} \right)\)
56. Find the coordinates of the point on a circle with radius 8 corresponding to an angle of \(\frac{7\pi}{4}.\)
57. Find the coordinates of the point on a circle with radius 16 corresponding to an angle of \(\frac{5\pi}{9}.\)
Solution (click to reveal)
\(\left( {–2.778,\ 15.757} \right)\)
58. State the domain of the sine and cosine functions.
59. State the range of the sine and cosine functions.
Solution (click to reveal)
\(\left\lbrack {–1,\ 1} \right\rbrack\)
Graphical
For the following exercises, use the given point on the unit circle to find the value of the sine and cosine of \(t.\)
60.
61.
Solution (click to reveal)
\(\sin t = \frac{1}{2},\cos t = - \frac{\sqrt{3}}{2}\)
62.
63.
Solution (click to reveal)
\(\sin t = - \frac{\sqrt{2}}{2},\cos t = - \frac{\sqrt{2}}{2}\)
64.
65.
Solution (click to reveal)
\(\sin t = \frac{\sqrt{3}}{2},\cos t = - \frac{1}{2}\)
66.
67.
Solution (click to reveal)
\(\sin t = - \frac{\sqrt{2}}{2},\cos t = \frac{\sqrt{2}}{2}\)
68.
69.
Solution (click to reveal)
\(\sin t = 0,\;\cos t = - 1\)
70.
71.
Solution (click to reveal)
\(\sin t = - 0.596,\;\cos t = 0.803\)
72.
73.
Solution (click to reveal)
\(\sin t = \frac{1}{2},\cos t = \frac{\sqrt{3}}{2}\)
74.
75.
Solution (click to reveal)
\(\sin t = - \frac{1}{2},\cos t = \frac{\sqrt{3}}{2}\)
76.
77.
Solution (click to reveal)
\(\sin t = 0.761,\cos t = - 0.649\)
78.
79.
Solution (click to reveal)
\(\sin t = 1,\cos t = 0\)
Technology
For the following exercises, use a graphing calculator to evaluate.
80. \(\sin\;\frac{5\pi}{9}\)
81. \(\cos\;\frac{5\pi}{9}\)
Solution (click to reveal)
−0.1736
82. \(\sin\;\frac{\pi}{10}\)
83. \(\cos\;\frac{\pi}{10}\)
Solution (click to reveal)
0.9511
84. \(\sin\;\frac{3\pi}{4}\)
85. \(\cos\;\frac{3\pi}{4}\)
Solution (click to reveal)
−0.7071
86. \(\sin\; 98{^\circ}\)
87. \(\cos\; 98{^\circ}\)
Solution (click to reveal)
−0.1392
88. \(\cos\; 310{^\circ}\)
89. \(\sin\; 310{^\circ}\)
Solution (click to reveal)
−0.7660
Extensions
For the following exercises, evaluate.
90. \(\sin\left( \frac{11\pi}{3} \right)\cos\left( \frac{- 5\pi}{6} \right)\)
91. \(\sin\left( \frac{3\pi}{4} \right)\cos\left( \frac{5\pi}{3} \right)\)
Solution (click to reveal)
\(\frac{\sqrt{2}}{4}\)
92. \(\sin\left( {- \frac{4\pi}{3}} \right)\cos\left( \frac{\pi}{2} \right)\)
93. \(\sin\left( \frac{- 9\pi}{4} \right)\cos\left( \frac{- \pi}{6} \right)\)
Solution (click to reveal)
\(- \frac{\sqrt{6}}{4}\)
94. \(\sin\left( \frac{\pi}{6} \right)\cos\left( \frac{- \pi}{3} \right)\)
95. \(\sin\left( \frac{7\pi}{4} \right)\cos\left( \frac{- 2\pi}{3} \right)\)
Solution (click to reveal)
\(\frac{\sqrt{2}}{4}\)
96. \(\cos\left( \frac{5\pi}{6} \right)\cos\left( \frac{2\pi}{3} \right)\)
97. \(\cos\left( \frac{- \pi}{3} \right)\cos\left( \frac{\pi}{4} \right)\)
Solution (click to reveal)
\(\frac{\sqrt{2}}{4}\)
98. \(\sin\left( \frac{- 5\pi}{4} \right)\sin\left( \frac{11\pi}{6} \right)\)
99. \(\sin(\pi)\sin\left( \frac{\pi}{6} \right)\)
Solution (click to reveal)
0
Real-World Applications
For the following exercises, use this scenario: A child enters a carousel that takes one minute to revolve once around. The child enters at the point \((0,1),\) that is, on the due north position. Assume the carousel revolves counter clockwise.
100. What are the coordinates of the child after 45 seconds?
101. What are the coordinates of the child after 90 seconds?
Solution (click to reveal)
\(\left( {0,–1} \right)\)
102. What are the coordinates of the child after 125 seconds?
103. When will the child have coordinates \((0.707,–0.707)\) if the ride lasts 6 minutes? (There are multiple answers.)
Solution (click to reveal)
37.5 seconds, 97.5 seconds, 157.5 seconds, 217.5 seconds, 277.5 seconds, 337.5 seconds
104. When will the child have coordinates \(\left( {–0.866,–0.5} \right)\) if the ride lasts 6 minutes?