Solutions to Exercises, Section 5.1

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Solutions to Exercises, Section 5.1

1. Find all numbers t such that (¹₃, t) is a point on the unit circle.

solution For (¹₃, t) to be a point on the unit circle means that the sum of the squares of the coordinates equals 1. In other words,

₁

3

2

+ t²= 1.

This simpliﬁes to the equation t² = ⁸₉, which implies that t = ^√₃⁸ or t = −^√₃⁸. Because√

8=√

4· 2 =√ 4·√

2= 2√

2, we can rewrite this as t= ²^√₃² or t= −²^√₃².

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3. Find all numbers t such that (t,−²₅) is a point on the unit circle.

solution For (t,−²₅) to be a point on the unit circle means that the sum of the squares of the coordinates equals 1. In other words,

t²+

−²₅2

= 1.

This simpliﬁes to the equation t² = ²¹₂₅, which implies that t = ^√₅²¹ or t= −^√₅²¹.

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5. Find the points where the line through the origin with slope 3 intersects the unit circle.

solution The line through the origin with slope 3 is characterized by the equation y= 3x. Substituting this value for y into the equation for the unit circle (x²+ y²= 1) gives

x²+ (3x)²= 1,

which simpliﬁes to the equation 10x² = 1. Thus x = ^√₁₀¹⁰ or x= −^√₁₀¹⁰. Using each of these values of x along with the equation y= 3x gives the points^√₁₀

10 ,³

√10 10

and

−^√₁₀¹⁰,−³^√₁₀¹⁰

as the points of intersection of the line y = 3x and the unit circle.

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7. Suppose an ant walks counterclockwise on the unit circle from the point (1, 0) to the endpoint of the radius that forms an angle of 70^◦with the positive horizontal axis. How far has the ant walked?

solution We need to ﬁnd the length of the circular arc on the unit circle corresponding to a 70^◦angle. This length equals ^70π₁₈₀, which equals ^7π₁₈.

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9. What angle corresponds to a circular arc on the unit circle with length^π₅? solution Let θ be such that the angle of θ degrees corresponds to an arc on the unit circle with length^π₅. Thus ₁₈₀^θπ = ^π₅. Solving this equation for θ, we get θ= 36. Thus the angle in question is 36^◦.

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11. What angle corresponds to a circular arc on the unit circle with length

5 2?

solution Let θ be such that the angle of θ degrees corresponds to an arc on the unit circle with length⁵₂. Thus₁₈₀^θπ = ⁵₂. Solving this equation for θ, we get θ= ⁴⁵⁰_π . Thus the angle in question is⁴⁵⁰_π ^◦, which is approximately equal to 143.2^◦.

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13. Find the lengths of both circular arcs on the unit circle connecting the points (1, 0) and^√₂

2 ,

√2 2

.

solution The radius of the unit circle ending at the point^√₂

2 ,

√2 2

makes an angle of 45^◦with the positive horizontal axis. One of the cir- cular arcs connecting (1, 0) and^√₂

2 ,

√2 2

is shown below as the thickened circular arc; the other circular arc connecting (1, 0) and ^√₂

2 ,

√2 2

is the unthickened part of the unit circle below.

The length of the thickened arc below is ^45π₁₈₀, which equals ^π₄. The entire unit circle has length 2π . Thus the length of the other circular arc below is 2π−^π₄, which equals ^7π₄ .

45

1

The thickened circular arc has length ^π₄. The other circular arc has length ^7π₄ .

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For each of the angles in Exercises 15–20, ﬁnd the endpoint of the radius of the unit circle that makes the given angle with the positive horizontal axis.

15. 120^◦

solution The radius making a 120^◦angle with the positive horizontal axis is shown below. The angle from this radius to the negative horizontal axis equals 180^◦− 120^◦, which equals 60^◦as shown in the ﬁgure below. Drop a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle as shown below. We already know that one angle of this right triangle is 60^◦; thus the other angle must be 30^◦, as labeled below:

120

60

30

1

The side of the right triangle opposite the 30^◦ angle has length ¹₂; the side of the right triangle opposite the 60^◦angle has length

√3

2 . Looking at the ﬁgure above, we see that the ﬁrst coordinate of the endpoint of the

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radius is the negative of the length of the side opposite the 30^◦angle, and the second coordinate of the endpoint of the radius is the length of the side opposite the 60^◦ angle. Thus the endpoint of the radius is

−¹₂,

√3 2

.

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17. −30^◦

solution The radius making a−30^◦angle with the positive horizontal axis is shown below. Draw a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle as shown below. We already know that one angle of this right triangle is 30^◦; thus the other angle must be 60^◦, as labeled below.

√3

2 . Looking at the ﬁgure below, we see that the ﬁrst coordinate of the endpoint of the radius is the length of the side opposite the 60^◦angle, and the second coordinate of the endpoint of the radius is the negative of the length of the side opposite the 30^◦ angle. Thus the endpoint of the radius is

^√₃

2 ,−¹₂ .

60

30 1

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19. 390^◦

solution The radius making a 390^◦angle with the positive horizontal axis is obtained by starting at the horizontal axis, making one complete counterclockwise rotation, and then continuing for another 30^◦. The resulting radius is shown below. Drop a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle as shown below. We already know that one angle of this right triangle is 30^◦; thus the other angle must be 60^◦, as labeled below.

The side of the right triangle opposite the 30^◦angle has length¹₂; the side opposite the 60^◦angle has length

√3

2 . Looking at the ﬁgure below, we see that the ﬁrst coordinate of the endpoint of the radius is the length of the side opposite the 60^◦ angle, and the second coordinate of the endpoint of the radius is the length of the side opposite the 30^◦ angle. Thus the endpoint of the radius is^√₃

2 ,¹₂ .

60

30

1

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For Exercises 21–26, ﬁnd the angle the radius of the unit circle ending at the given point makes with the positive horizontal axis. Among the inﬁnitely many possible correct solutions, choose the one with the smallest absolute value.

21.

−¹₂,

√3 2

solution Draw the radius whose endpoint is

−¹₂,

√3 2

. Drop a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle. The hypotenuse of this right triangle is a radius of the unit circle and thus has length 1. The horizontal side has length ¹₂ and the vertical side of this triangle has length

√3

2 because the endpoint of the radius is

−¹₂,

√3 2

.

120

60

30

1

Thus we have a 30^◦- 60^◦- 90^◦ triangle, with the 30^◦ angle opposite the horizontal side of length ¹₂, as labeled above. Because 180− 60 = 120,

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the radius makes a 120^◦angle with the positive horizontal axis, as shown above.

In addition to making a 120^◦angle with the positive horizontal axis, this radius also makes with the positive horizontal axis angles of 480^◦, 840^◦, and so on. This radius also makes with the positive horizontal axis angles of−240^◦,−600^◦, and so on. But of all the possible choices for this angle, the one with the smallest absolute value is 120^◦.

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23. ^√₂

2 ,−^√₂²

solution Draw the radius whose endpoint is^√₂

2 ,−^√₂²

. Draw a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle. The hypotenuse of this right triangle is a radius of the unit circle and thus has length 1. The horizontal side has length

√2

2 and the vertical side of this triangle also has length

√2

2 because the endpoint of the radius is^√₂

2 ,−^√₂² .

45

45 1

Thus we have here an isosceles right triangle, with two angles of 45^◦as labeled above.

In addition to making a −45^◦ angle with the positive horizontal axis, this radius also makes with the positive horizontal axis angles of 315^◦, 675^◦, and so on. This radius also makes with the positive horizontal axis angles of−405^◦,−765^◦, and so on. But of all the possible choices for this angle, the one with the smallest absolute value is−45^◦.

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25.

−^√₂²,−^√₂²

solution Draw the radius whose endpoint is

−^√₂²,−^√₂²

. Draw a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle. The hypotenuse of this right triangle is a radius of the unit circle and thus has length 1. The horizontal side has length

√2

2 and the vertical side of this triangle also has length

√2

2 because the endpoint of the radius is

−^√₂²,−^√₂² .

45

45 135

1

Thus we have here an isosceles right triangle, with two angles of 45^◦as labeled above. Because the radius makes a 45^◦angle with the negative horizontal axis, it makes a−135^◦angle with the positive horizontal axis, as shown below (because 135^◦= 180^◦− 45^◦).

In addition to making a −135^◦ angle with the positive horizontal axis, this radius also makes with the positive horizontal axis angles of 225^◦, 585^◦, and so on. This radius also makes with the positive horizontal axis

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angles of−495^◦,−855^◦, and so on. But of all the possible choices for this angle, the one with the smallest absolute value is−135^◦.

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27. Find the lengths of both circular arcs on the unit circle connecting the point₁

2,

√3 2

and the point that makes an angle of 130^◦with the positive horizontal axis.

solution The radius of the unit circle ending at the point₁

2,

√3 2

makes an angle of 60^◦with the positive horizontal axis. One of the circular arcs connecting₁

2,

√3 2

and the point that makes an angle of 130^◦ with the positive horizontal axis is shown below as the thickened circular arc; the other circular arc connecting these two points is the unthickened part of the unit circle below.

The thickened arc below corresponds to an angle of 70^◦(because 70^◦= 130^◦− 60^◦). Thus the length of the thickened arc below is ^70π₁₈₀, which equals ^7π₁₈. The entire unit circle has length 2π . Thus the length of the other circular arc below is 2π− ^7π₁₈, which equals ^29π₁₈ .

60 130

1

The thickened circular arc has length ^7π₁₈. The other circular arc has length ^29π₁₈ .

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29. Find the lengths of both circular arcs on the unit circle connecting the point

−^√₂²,−^√₂²

and the point that makes an angle of 125^◦ with the positive horizontal axis.

solution The radius of the unit circle ending at

−^√₂²,−^√₂²

makes an angle of 225^◦with the positive horizontal axis (because 225^◦= 180^◦+45^◦).

One of the circular arcs connecting

−^√₂²,−^√₂²

and the point that makes an angle of 125^◦with the positive horizontal axis is shown below as the thickened circular arc; the other circular arc connecting these two points is the unthickened part of the unit circle below.

The thickened arc below corresponds to an angle of 100^◦(because 100^◦= 225^◦− 125^◦). Thus the length of the thickened arc below is ^100π₁₈₀ , which equals ^5π₉ . The entire unit circle has length 2π . Thus the length of the other circular arc below is 2π− ^5π₉ , which equals ^13π₉ .

125 225

1

The thickened circular arc has length ^5π₉ . The other circular arc has length ^13π₉ .

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31. What is the slope of the radius of the unit circle that has a 30^◦angle with the positive horizontal axis?

solution The radius of the unit circle that has a 30^◦ angle with the positive horizontal axis has its initial point at (0, 0) and its endpoint at

^√₃

2 ,¹₂

. Thus the slope of this radius is

1 2− 0

√3 2 − 0 , which equals ^√¹₃, which equals

√3 3 .

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In Exercises 1–8, convert each angle to radians.

1. 15^◦

solution Start with the equation

360^◦= 2π radians.

Divide both sides by 360 to obtain 1^◦= π

180 radians.

Now multiply both sides by 15, obtaining

15^◦= 15π

180 radians= π

12 radians.

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3. −45^◦

180 radians.

Now multiply both sides by−45, obtaining

−45^◦= −45π

180 radians= −π

4 radians.

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5. 270^◦

180 radians.

270^◦= 270π

180 radians= 3π

2 radians.

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7. 1080^◦

180 radians.

1080^◦= 1080π

180 radians= 6π radians.

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In Exercises 9–16, convert each angle to degrees.

9. 4π radians

2π radians= 360^◦. Multiply both sides by 2, obtaining

4π radians= 2 · 360^◦= 720^◦.

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11. ^π₉ radians

2π radians= 360^◦. Divide both sides by 2 to obtain

π radians= 180^◦. Now divide both sides by 9, obtaining

π

9 radians= 180 9

◦

= 20^◦.

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13. 3 radians

2π radians= 360^◦. Divide both sides by 2π to obtain

1 radian= 180 π

◦

. Now multiply both sides by 3, obtaining

3 radians= 3 ·180 π

◦= 540 π

◦

.

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15. −^2π₃ radians

2π radians= 360^◦. Divide both sides by 2 to obtain

π radians= 180^◦. Now multiply both sides by−²₃, obtaining

−2π

3 radians= −2

3· 180^◦= −120^◦.

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17. Suppose an ant walks counterclockwise on the unit circle from the point (0, 1) to the endpoint of the radius that forms an angle of ^5π₄ radians with the positive horizontal axis. How far has the ant walked?

solution The radius whose endpoint equals (0, 1) makes an angle of

π

2 radians with the positive horizontal axis. This radius corresponds to the smaller angle shown below.

Because ^5π₄ = π +^π₄, the radius that forms an angle of ^5π₄ radians with the positive horizontal axis lies ^π₄ radians beyond the negative horizontal axis (half-way between the negative horizontal axis and the negative vertical axis). Thus the ant ends its walk at the endpoint of the radius corresponding to the larger angle shown below:

1

The ant walks along the thickened circular arc shown above. This circular arc corresponds to an angle of^5π₄ −^π₂ radians, which equals ^3π₄ radians.

Thus the distance walked by the ant is ^3π₄ .

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19. Find the lengths of both circular arcs of the unit circle connecting the point (1, 0) and the endpoint of the radius that makes an angle of 3 radians with the positive horizontal axis.

solution Because 3 is a bit less than π , the radius that makes an angle of 3 radians with the positive horizontal axis lies a bit above the negative horizontal axis, as shown below. The thickened circular arc corresponds to an angle of 3 radians and thus has length 3. The entire unit circle has length 2π . Thus the length of the other circular arc is 2π− 3, which is approximately 3.28.

1

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21. Find the lengths of both circular arcs of the unit circle connecting the point^√₂

2 ,−^√₂²

and the point whose radius makes an angle of 1 radian with the positive horizontal axis.

solution The radius of the unit circle whose endpoint equals^√₂

2 ,−^√₂² makes an angle of−^π₄ radians with the positive horizontal axis, as shown with the clockwise arrow below. The radius that makes an angle of 1 radian with the positive horizontal axis is shown with a counterclockwise arrow.

1

Thus the thickened circular arc above corresponds to an angle of 1+^π₄ and thus has length 1+^π₄, which is approximately 1.79. The entire unit circle has length 2π . Thus the length of the other circular arc below is 2π− (1 +^π₄), which equals ^7π₄ − 1, which is approximately 4.50.

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23. For a 16-inch pizza, ﬁnd the area of a slice with angle ³₄ radians.

solution Pizzas are measured by their diameters; thus this pizza has a radius of 8 inches. Thus the area of the slice is ¹₂·³₄· 8², which equals 24 square inches.

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25. Suppose a slice of a 12-inch pizza has an area of 20 square inches. What is the angle of this slice?

solution This pizza has a radius of 6 inches. Let θ denote the angle of this slice, measured in radians. Then

20= ¹₂θ· 6².

Solving this equation for θ, we get θ= ¹⁰₉ radians.

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27. Suppose a slice of pizza with an angle of⁵₆ radians has an area of 21 square inches. What is the diameter of this pizza?

solution Let r denote the radius of this pizza. Thus

21= ¹₂· ⁵₆r². Solving this equation for r , we get r =

252

5 ≈ 7.1. Thus the diameter of the pizza is approximately 14.2 inches.

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For each of the angles in Exercises 29–34, ﬁnd the endpoint of the radius of the unit circle that makes the given angle with the positive horizontal axis.

29. ^5π₆ radians

solution For this exercise it may be easier to convert to degrees. Thus we translate ^5π₆ radians to 150^◦.

The radius making a 150^◦angle with the positive horizontal axis is shown below. The angle from this radius to the negative horizontal axis equals 180^◦− 150^◦, which equals 30^◦ as shown in the ﬁgure below. Drop a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle as shown below. We already know that one angle of this right triangle is 30^◦; thus the other angle must be 60^◦, as labeled below.

√3

2 . Looking at the ﬁgure below, we see that the ﬁrst coordinate of the endpoint of the radius is the negative of the length of the side opposite the 60^◦angle, and the second coordinate of the endpoint of the radius is the length of the side opposite the 30^◦ angle. Thus the endpoint of the radius is

−^√₂³,¹₂ .

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150

30

60

1

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31. −^π₄ radians

solution For this exercise it may be easier to convert to degrees. Thus we translate−^π₄ radians to−45^◦.

The radius making a −45^◦ angle with the positive horizontal axis is shown below. Draw a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle as shown below. We already know that one angle of this right triangle is 45^◦; thus the other angle must also be 45^◦, as labeled below.

The hypotenuse of this right triangle is a radius of the unit circle and thus has length 1. The other two sides each have length

√2

2 . Looking at the ﬁgure below, we see that the ﬁrst coordinate of the endpoint of the radius is

√2

2 and the second coordinate of the endpoint of the radius is

−^√₂². Thus the endpoint of the radius is^√₂

2 ,−^√₂² .

45

45 1

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33. ^5π₂ radians

solution Note that ^5π₂ = 2π +^π₂. Thus the radius making an angle of

5π

2 radians with the positive horizontal axis is obtained by starting at the horizontal axis, making one complete counterclockwise rotation (which is 2π radians), and then continuing for another ^π₂ radians. The resulting radius is shown below. Its endpoint is (0, 1).

1

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Give exact values for the quantities in Exercises 1–10. Do not use a cal- culator for any of these exercises—otherwise you will likely get decimal approximations for some solutions rather than exact answers. More im- portantly, good understanding will come from working these exercises by hand.

1. (a) cos 3π (b) sin 3π

solution Because 3π = 2π + π, an angle of 3π radians (as measured counterclockwise from the positive horizontal axis) consists of a com- plete revolution around the circle (2π radians) followed by another π radians (180^◦), as shown below. The endpoint of the corresponding ra- dius is (−1, 0). Thus cos 3π = −1 and sin 3π = 0.

1

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3. (a) cos^11π₄ (b) sin^11π₄

solution Because ^11π₄ = 2π + ^π₂ + ^π₄, an angle of ^11π₄ radians (as measured counterclockwise from the positive horizontal axis) consists of a complete revolution around the circle (2π radians) followed by an- other ^π₂ radians (90^◦), followed by another ^π₄ radians (45^◦), as shown below. Hence the endpoint of the corresponding radius is

−^√₂²,

√2 2

. Thus cos^11π₄ = −^√₂² and sin^11π₄ = ^√₂².

1

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5. (a) cos^2π₃ (b) sin^2π₃

solution Because ^2π₃ = ^π₂ + ^π₆, an angle of ^2π₃ radians (as measured counterclockwise from the positive horizontal axis) consists of^π₂ radians (90^◦radians) followed by another ^π₆ radians (30^◦), as shown below. The endpoint of the corresponding radius is

−¹₂,

√3 2

. Thus cos^2π₃ = −¹₂and sin^2π₃ = ^√₂³.

1

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7. (a) cos 210^◦ (b) sin 210^◦

solution Because 210= 180+30, an angle of 210^◦(as measured counterclockwise from the positive horizontal axis) consists of 180^◦followed by another 30^◦, as shown below. The endpoint of the corresponding radius is

−^√₂³,−¹₂

. Thus cos 210^◦= −^√₂³ and sin 210^◦= −¹₂.

1

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9. (a) cos 360045^◦ (b) sin 360045^◦

solution Because 360045= 360 × 1000 + 45, an angle of 360045^◦(as measured counterclockwise from the positive horizontal axis) consists of 1000 complete revolutions around the circle followed by another 45^◦. The endpoint of the corresponding radius is^√₂

2 ,

√2 2

. Thus

cos 360045^◦= ^√₂² and sin 360045^◦= ^√₂².

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11. Find the smallest number θ larger than 4π such that cos θ= 0.

solution Note that

0= cos^π₂ = cos^3π₂ = cos^5π₂ = . . .

and that the only numbers whose cosine equals 0 are of the form^(2n+1)π₂ , where n is an integer. The smallest number of this form larger than 4π is

9π

2 . Thus ^9π₂ is the smallest number larger than 4π whose cosine equals 0.

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13. Find the four smallest positive numbers θ such that cos θ= 0.

solution Think of a radius of the unit circle whose endpoint is (1, 0). If this radius moves counterclockwise, forming an angle of θ with the pos- itive horizontal axis, the ﬁrst coordinate of its endpoint ﬁrst becomes 0 when θ equals^π₂ (which equals 90^◦), then again when θ equals^3π₂ (which equals 270^◦), then again when θ equals ^5π₂ (which equals 360^◦+ 90^◦, or 450^◦), then again when θ equals ^7π₂ (which equals 360^◦+ 270^◦, or 630^◦), and so on. Thus the four smallest positive numbers θ such that cos θ= 0 are ^π₂,^3π₂ , ^5π₂ , and ^7π₂ .

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15. Find the four smallest positive numbers θ such that sin θ= 1.

solution Think of a radius of the unit circle whose endpoint is (1, 0).

If this radius moves counterclockwise, forming an angle of θ with the positive horizontal axis, then the second coordinate of its endpoint ﬁrst becomes 1 when θ equals^π₂ (which equals 90^◦), then again when θ equals

5π

2 (which equals 360^◦+90^◦, or 450^◦), then again when θ equals^9π₂ (which equals 2× 360^◦+ 90^◦, or 810^◦), then again when θ equals ^13π₂ (which equals 3× 360^◦+ 90^◦, or 1170^◦), and so on. Thus the four smallest positive numbers θ such that sin θ= 1 are^π₂, ^5π₂ , ^9π₂ , and ^13π₂ .

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17. Find the four smallest positive numbers θ such that cos θ= −1.

If this radius moves counterclockwise, forming an angle of θ with the positive horizontal axis, the ﬁrst coordinate of its endpoint ﬁrst becomes

−1 when θ equals π (which equals 180^◦), then again when θ equals 3π (which equals 360^◦+180^◦, or 540^◦), then again when θ equals 5π (which equals 2× 360^◦+ 180^◦, or 900^◦), then again when θ equals 7π (which equals 3× 360^◦+ 180^◦, or 1260^◦), and so on. Thus the four smallest positive numbers θ such that cos θ= −1 are π, 3π, 5π, and 7π.

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19. Find the four smallest positive numbers θ such that sin θ= ¹₂.

solution Think of a radius of the unit circle whose endpoint is (1, 0). If this radius moves counterclockwise, forming an angle of θ with the posi- tive horizontal axis, the second coordinate of its endpoint ﬁrst becomes¹₂ when θ equals^π₆ (which equals 30^◦), then again when θ equals^5π₆ (which equals 150^◦), then again when θ equals ^13π₆ (which equals 360^◦+ 30^◦, or 390^◦), then again when θ equals ^17π₆ (which equals 360^◦+ 150^◦, or 510^◦), and so on. Thus the four smallest positive numbers θ such that sin θ= ¹₂ are ^π₆,^5π₆ , ^13π₆ , and ^17π₆ .

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21. Suppose 0 < θ < ^π₂ and cos θ= ²₅. Evaluate sin θ.

solution We know that

(cos θ)²+ (sin θ)²= 1.

Thus

(sin θ)²= 1 − (cos θ)²

= 1 −2 5

2

= 21 25.

Because 0 < θ < ^π₂, we know that sin θ > 0. Thus taking square roots of both sides of the equation above gives

sin θ=

√21 5 .

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23. Suppose ^π₂ < θ < π and sin θ= ²₉. Evaluate cos θ.

(cos θ)²+ (sin θ)²= 1.

Thus

(cos θ)²= 1 − (sin θ)²

= 1 −2 9

2

= 77 81.

Because ^π₂ < θ < π , we know that cos θ < 0. Thus taking square roots of both sides of the equation above gives

cos θ= −

√77 9 .

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25. Suppose−^π₂ < θ < 0 and cos θ= 0.1. Evaluate sin θ.

(cos θ)²+ (sin θ)²= 1.

Thus

(sin θ)²= 1 − (cos θ)²

= 1 − (0.1)²

= 0.99.

Because−^π₂ < θ < 0, we know that sin θ < 0. Thus taking square roots of both sides of the equation above gives

sin θ= −√

0.99≈ −0.995.

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27. Find the smallest number x such that sin(e^x)= 0.

solution Note that e^x is an increasing function. Because e^xis positive for every real number x, and because π is the smallest positive number whose sine equals 0, we want to choose x so that e^x = π. Thus x = ln π.

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29. Find the smallest positive number x such that sin(x²+ x + 4) = 0.

solution Note that x²+ x + 4 is an increasing function on the interval [0,∞). If x is positive, then x²+ x + 4 > 4. Because 4 is larger than π but less than 2π , the smallest number bigger than 4 whose sine equals 0 is 2π . Thus we want to choose x so that x²+ x + 4 = 2π. In other words, we need to solve the equation

x²+ x + (4 − 2π) = 0.

Using the quadratic formula, we see that the solutions to this equation are

x= −1 ±√

8π− 15

2 .

A calculator shows that choosing the plus sign in the equation above gives x ≈ 1.0916 and choosing the minus sign gives x ≈ −2.0916. We seek only positive values of x, and thus we choose the plus sign in the equation above, getting x≈ 1.0916.

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1. Find the four smallest positive numbers θ such that tan θ= 1.

If this radius moves counterclockwise, forming an angle of θ with the positive horizontal axis, then the ﬁrst and second coordinates of its end- point ﬁrst become equal (which is equivalent to having tan θ= 1) when θ equals^π₄ (which equals 45^◦), then again when θ equals ^5π₄ (which equals 225^◦), then again when θ equals ^9π₄ (which equals 360^◦+ 45^◦, or 405^◦), then again when θ equals ^13π₄ (which equals 360^◦+ 225^◦, or 585^◦), and so on.

Thus the four smallest positive numbers θ such that tan θ= 1 are ^π₄,^5π₄ ,

9π

4 , and ^13π₄ .

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3. Suppose 0 < θ <^π₂ and cos θ= ¹₅. Evaluate:

(a) sin θ (b) tan θ

solution The ﬁgure below gives a sketch of the angle involved in this exercise:

Θ

1

The angle between 0 and ^π₂ whose cosine equals ¹₅.

(a) We know that

(cos θ)²+ (sin θ)²= 1.

Thus₁

5

2

+ (sin θ)²= 1. Solving this equation for (sin θ)²gives

(sin θ)²= 24 25.

The sketch above shows that sin θ > 0. Thus taking square roots of both sides of the equation above gives

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sin θ=

√24

5 =

√4· 6 5 = 2√

6 5 . (b)

tan θ= sin θ cos θ =

2√ 6 5 1 5

= 2√ 6.

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5. Suppose^π₂ < θ < π and sin θ= ²₃. Evaluate:

(a) cos θ (b) tan θ

Θ

1

The angle between ^π₂ and π whose sine equals ²₃.

(a) We know that

(cos θ)²+ (sin θ)²= 1.

Thus (cos θ)²+₂

3

2

= 1. Solving this equation for (cos θ)²gives

(cos θ)²= 5 9.

The sketch above shows that cos θ < 0. Thus taking square roots of both sides of the equation above gives

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cos θ= −

√5 3 . (b)

tan θ= sin θ cos θ = −

2

√3 5 3

= − 2

√5= −2√ 5 5 .

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7. Suppose−^π₂ < θ < 0 and cos θ= ⁴₅. Evaluate:

(a) sin θ (b) tan θ

Θ ¹

The angle between−^π₂ and 0 whose cosine equals ⁴₅.

(a) We know that

(cos θ)²+ (sin θ)²= 1.

Thus₄

5

2

+ (sin θ)²= 1. Solving this equation for (sin θ)²gives

(sin θ)²= 9 25.

The sketch above shows that sin θ < 0. Thus taking square roots of both sides of the equation above gives

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sin θ= −3 5. (b)

tan θ= sin θ cos θ = −

3 5 4 5

= −3 4.

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9. Suppose 0 < θ <^π₂ and tan θ= ¹₄. Evaluate:

(a) cos θ (b) sin θ

Θ 1

The angle between 0 and ^π₂ whose tangent equals ¹₄.

(a) Rewrite the equation tan θ = ¹₄ in the form _{cos θ}^{sin θ} = ¹₄. Multiplying both sides of this equation by cos θ, we get

sin θ= ¹₄cos θ.

Substitute this expression for sin θ into the equation (cos θ)²+(sin θ)²= 1, getting

(cos θ)²+₁₆¹(cos θ)²= 1, which is equivalent to

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(cos θ)²= 16 17.

The sketch above shows that cos θ > 0. Thus taking square roots of both sides of the equation above gives

cos θ= 4

√17 = 4√ 17 17 . (b) We have already noted that sin θ= ¹₄cos θ. Thus

sin θ=

√17 17 .

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11. Suppose−^π₂ < θ < 0 and tan θ= −3. Evaluate:

(a) cos θ (b) sin θ

Θ ¹

The angle between−^π₂ and 0 whose tangent equals−3.

(a) Rewrite the equation tan θ= −3 in the form_{cos θ}^{sin θ} = −3. Multiplying both sides of this equation by cos θ, we get

sin θ= −3 cos θ.

Substitute this expression for sin θ into the equation (cos θ)²+(sin θ)²= 1, getting

(cos θ)²+ 9(cos θ)²= 1, which is equivalent to

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(cos θ)²= 1 10.

The sketch above shows that cos θ > 0. Thus taking square roots of both sides of the equation above gives

cos θ= 1

√10 =

√10 10 . (b) We have already noted that sin θ= −3 cos θ. Thus

sin θ= −3√ 10 10 .

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Given that

cos 15^◦ =

2+√ 3

2 and sin 22.5^◦ =

2−√ 2

2 ,

in Exercises 13–22 ﬁnd exact expressions for the indicated quantities.

[These values for cos 15^◦and sin 22.5^◦will be derived in Examples 4 and 5 in Section 6.3.]

13. sin 15^◦

(cos 15^◦)²+ (sin 15^◦)²= 1.

Thus

(sin 15^◦)²= 1 − (cos 15^◦)²

= 1 −2 + √3 2

2

= 1 −2+√ 3 4

= 2−√ 3 4 .

Because sin 15^◦ > 0, taking square roots of both sides of the equation above gives

sin 15^◦=

2−√ 3

2 .

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15. tan 15^◦ solution

tan 15^◦= sin 15^◦ cos 15^◦

=

2−√ 3

2+√ 3

=

2−√ 3

2+√ 3·

2−√ 3

= 2−√

√ 3 4− 3

= 2 −√ 3

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17. cot 15^◦ solution

cot 15^◦= 1 tan 15^◦

= 1

2−√ 3

= 1

2−√

3·2+√ 3 2+√

3

= 2+√ 3 4− 3

= 2 +√ 3

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19. csc 15^◦ solution

csc 15^◦= 1 sin 15^◦

= 2

2−√ 3

= 2

2−√ 3·

2+√ 3

= 2 2+√

√ 3 4− 3

= 2

2+√ 3

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21. sec 15^◦ solution

sec 15^◦= 1 cos 15^◦

= 2

2+√ 3

= 2

2+√ 3·

2−√ 3

= 2 2−√

√ 3 4− 3

= 2

2−√ 3

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Suppose u and ν are in the interval (0,^π₂), with tan u = 2 and tan ν = 3.

In Exercises 23–32, ﬁnd exact expressions for the indicated quantities.

23. cot u solution

cot u= 1 tan u

= 1 2

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25. cos u

2= tan u

= sin u cos u. To ﬁnd cos u, make the substitution sin u=

1− (cos u)² in the equa- tion above (this substitution is valid because we know that 0 < u < ^π₂ and thus sin u > 0), getting

2=

1− (cos u)² cos u .

Now square both sides of the equation above, then multiply both sides by (cos u)² and rearrange to get the equation

5(cos u)²= 1.

Because 0 < u < ^π₂, we see that cos u > 0. Thus taking square roots of both sides of the equation above gives cos u= ^√¹₅, which can be rewritten as cos u= ^√₅⁵.

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27. sin u solution

sin u=

1− (cos u)²

=

1−1 5

=

4 5

= 2

√5

= 2√ 5 5

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29. csc u solution

csc u= 1 sin u

=

√5 2

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31. sec u solution

sec u= 1 cos u

=√ 5

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33. Find the smallest number x such that tan e^x= 0.

solution Note that e^x is an increasing function. Because e^xis positive for every real number x, and because π is the smallest positive number whose tangent equals 0, we want to choose x so that e^x = π. Thus x= ln π ≈ 1.14473.

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Use the right triangle below for Exercises 1–76. This triangle is not drawn to scale corresponding to the data in the exercises.

u

Ν

a

b c

1. Suppose a= 2 and b = 7. Evaluate c.

solution The Pythagorean Theorem implies that c²= 2²+ 7². Thus c=

2²+ 7²=√ 53.

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3. Suppose a= 2 and b = 7. Evaluate cos u.

solution

cos u= adjacent side hypotenuse = a

c = 2

√53 = 2√ 53 53

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5. Suppose a= 2 and b = 7. Evaluate sin u.

solution

sin u= opposite side hypotenuse = b

c = 7

√53 = 7√ 53 53

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7. Suppose a= 2 and b = 7. Evaluate tan u.

solution

tan u= opposite side adjacent side = b

a = 7 2

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9. Suppose a= 2 and b = 7. Evaluate cos ν.

solution

cos ν= adjacent side hypotenuse = b

c = 7

√53 = 7√ 53 53

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11. Suppose a= 2 and b = 7. Evaluate sin ν.

solution

sin ν= opposite side hypotenuse = a

c = 2

√53 = 2√ 53 53

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13. Suppose a= 2 and b = 7. Evaluate tan ν.

solution

tan ν= opposite side adjacent side = a

b = 2 7

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15. Suppose b= 2 and c = 7. Evaluate a.

solution The Pythagorean Theorem implies that a²+ 2²= 7². Thus a=

7²− 2²=√ 45=√

9· 5 =√ 9·√

5= 3√ 5.