BrownMath.com → Trig w/o Tears → Problem Solutions
Updated 6 Nov 2020

Trig without Tears:

# Solutions to Practice Problems

## Solutions for Part 1: Introduction

1 Find these angles in degrees: (a) π/6;   (b) 2π;   (c) 1 (that’s right, radian angles aren’t necessarily fractions or multiples of π).

Solutions:

(a) (π/6) × (180°/π) = 30°

(b) 2π × (180°/π) = 360°

(c) 1 × (180°/π) = (180/π)° ≈ 57.3°

2 Which is the correct definition of an acute angle, in interval notation?
(a) (0°, 90°)   (b) [0°, 90°]

Answer: (0°, 90°) is 0 to 90 degrees excluding 0° and 90°; [0°, 90°] is 0 to 90 degrees including 0° and 90°. Acute angles are between 0° and 90° exclusive, so the answer is (a) (0°, 90°).

3 Two angles of a triangle are 80° and 40°. Fine the third angle.

Solution: The inside angles of a triangle must always add to 180°. 80° + 40° = 120°, so to make the full 180° the third angle must be 60°.

4 A triangle has an angle of 90°. The two short sides (next to that angle) are 5 and 12. Find the third side.

Solution: Cue the Pythagorean Theorem!

c² = a² + b²

c² = 5² + 12²

c² = 25 + 144 = 169

c = √169 = 13

5 Find these angles in radian measure: (a) 60°   (b) 126°;   (c) 45°.

Where possible, give an exact answer rather than a decimal approximation.

Solutions:

(a) 60° + (π/180°) = π/3.

(b) 126° × (π/180°) ≈ 2.20

(c) 45° × (π/180°) = π/4

Notice that you don’t have to say “radians” when giving an angle in radian measure, though it wouldn’t be wrong. In this book, angles in degrees have the degree mark (°), so I’ll only say “radians” when it’s necessary to avoid confusion.

6 Who said, “The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side”? Is that correct?

Answer: It was the Scarecrow, in the movie The Wizard of Oz (1939). And no, it sounds mathy but it’s bosh. It can’t possibly be true for any triangle, isosceles or not. (Can you see why?)

7 On a circular clock face, which numbers are the boundaries of each quadrant?

## Solutions for Part 2: The Six Functions

1 Find all six functions of the angle 30°. Find sine, cosine, and tangent of 60°. Solution: First off, you need the length of the horizontal side. You remember the theorem of Pythagoras: 1² + b² = 2², from which you get b = √3. After that, it’s just a matter of remembering the definitions. If you need a refresher, you’ll find sine and cosine at equation 1, tangent at equation 4, and the others at equation 5.

sin 30° = 1/2

cos 30° = √3/2

tan 30° = 1/√3 or √3/3

cot 30° = 1/(1/√3) = √3

sec 30° = 1/(√3/2) = 2/√3 or (2 √3)/3

csc 30° = 1/(1/2) = 2

sin 60° = √3/2

cos 60° = 1/2

tan 60° = √3/1 = √3

Since 60° = 90° − 30°, notice that sin 60° = cos 30°, cos 60° = sin 30°, and tan 60° = cot 30°.

2 Find sin A, sin B, tan A, and tan B.

Solution:

sin A = 3/5 or 0.6

sin B = 4/5 or 0.8

tan A = 3/4 or 0.75

tan B = 4/3 ≈ 1.33

Incidentally, A ≈ 36.87°, and B ≈ 53.13°.

3 A ≈ 53.13°. Find the approximate area of the triangle. Hint: the area of a triangle is base × height/2.

Solution: You have the base (5), so you just need the height. But sin A = h/3, so h = 3 × sin A. The area therefore is (5 × 3 × sin A)/2 ≈ 5.99999.

## Solutions for Part 3: Functions of Special Angles

1 (a) Sketch a 45-45-90° triangle with hypotenuse of 1. Label the size of each angle and the exact length of each side, not a calculator approximation. (Hint: Since the two acute angles are equal, the two short sides must be equal. That and the Theorem of Pythagoras is enough to let you find them.)

(b) Draw a triangle with all sides equal to 1. If all sides are equal, all angles must be equal. Knowing that they add to 180°, fill in the value of each angle. Now drop a perpendicular from the top of the triangle to the middle of the opposite side. You now have two 30-60-90° triangles with hypotenuse of 1. Fill in the exact lengths of the short sides of those two triangles.

For the rest of these problems, refer to these sketches if you need to. Give exact answers, not decimal approximations.

Solution: Check your sketches against the ones in the chapter.

2 Find tan 45°, cos 45°, sin 90°, cos 30°, sin 30°, cos 90°.

Answers: tan 45° = 1; cos 45° = 1/√2 or √2/2; sin 90° = 1; cos 30° = √3/2; sin 30° = 1/2; cos 90° = 0.

3 Find sin(π/4), cos(π/6), tan(π/3).

Answers: sin(π/4) = √2/2 or 1/√2; cos(π/6) = √3/2; tan(π/3) = √3

4 Find each of the following angles from the clues, assuming all the angles are between 0 and π/2 (0° and 90°) inclusive. Give each answer in degrees and radians. sin A = 0;  cos B = √3/2;  sin C = 1/2;  sin D = 1;  tan E = 1;  cos F = 1/2;  tan G = 0;  tan H = √3;  cos I = 1;  cos J = 0.

sin A = 0 ⇒ A = 0° or 0 [radians]

cos B = √3/2 ⇒ B = 30° or π/6

sin C = 1/2 ⇒ C = 30° or π/6

sin D = 1 ⇒ D = 90° or π/2

tan E = 1 ⇒ E = 45° or π/4

cos F = 1/2 ⇒ F = 60° or π/3

tan G = 0 ⇒ G = 0° or 0 [radians]

tan H = √3H = 60° or π/3

cos I = 1 ⇒ I = 0° or 0 [radians]

cos J = 0 ⇒ J = 90° or π/2

5 Find sec 60° and cot 30°. Hint: Remember how the secant and cotangent are defined in terms of the “big three” functions sine, cosine, and tangent.

Solutions: sec 60° = 1/(cos 60°) = 1/(1/2) ⇒ sec 60° = 2

cot 30° = 1/(tan 30°) = 1/(√3/3) = 3/√3 = √3cot 30° = √3

You could also do the second one using equation 6:

cot 30° = tan(90°−30°) = tan 60° = √3

6 Find sin 120°, cos 120°, and tan 120°.

Solution:

sin 120° = sin(180° − 120°) = √3/2.

cos 120° = −cos(180° − 120°) = −1/2.

What about the tangent? You don’t have a supplement rule for it, but you have another way to find the answer:

tan 120° = sin 120° / cos 120° = (√3/2) / (−1/2) = −√3

7Even though you can always get the tangent of a supplementary angle from the sine and cosine, it’s a time-saver to have a rule for the supplement of a tangent. The last problem’s solution suggested what that rule might be.

Prove: tan(180° − A) = −tan A.

Proof:

tan(180° − A) = sin(180° − A) / cos(180° − A)

tan(180° − A) = sin A / (−cos A)

tan(180° − A) = −sin A / cos A

tan(180° − A) = −tan A QED

8 Find tan 150°.

Solution: tan 150° = −tan(180° − 150°) = −tan 30° = −√3/3

## Solutions for Part 4: Solving Triangles

1 You have a right triangle (C = 90°) with short sides a = 88 and b = 37. Solve the triangle.

Always start with a sketch. From the sketch, you can see right away that this is the SAS case, or side-angle-side. To get the third side, you need the Law of Cosines, equation 31.

c² = a² + b² − 2ab × cos C

c² = 37² + 88² − 2 × 37 × 88 × cos 90°

Notice that when the included angle C is 90°, cos C = 0 and you’ve just got the Pythagorean Theorem.

c² = 37² + 88²

c = √37² + 88²

c ≈ 95.5

To find the two angles, you could use the Law of Sines, but why not take advantage of the right angle and use the tangent?

tan A = 37/88 ⇒ A ≈ 22.8°

Of course tan B = 88/37, but you also know that

A + B = 90° ⇒ B = 90° − AB ≈ 67.2°

2(Sketch this problem as you read through it.) In a state park, a river flows virtually straight for 1800 m. You want to build a monorail from A, one end of this stretch, to a point C on the far shore. You also want to build a foot bridge from B, at the other end of this stretch of the river, to the same point C on the far shore. At A, the angle between your sight lines to B and C is 67°. At B, the angle between your sight lines to A and C is 38°.

How long must the monorail and the foot bridge be?

Bonus question: If the river has the same width all along the stretch from A to B, how wide is it? In the sketch, b is the monorail, a is the foot bridge, and w is the width of the river. Since you know two angles and the included side, this is the ASA case. Although you don’t really care about angle C, you have to find it so that you can use the Law of Sines to get sides a and b:

C = 180° − 67° − 38° = 75°

Obviously, when I eyeballed the angles A and B I didn’t estimate them very well! But that’s okay—the sketch is close enough to be useful.

How long is the foot bridge? From the Law of Sines,

a/sin A = c/sin C a = c sin A/sin C

a = 1800 × sin 67°/sin 75°

Foot bridge: a ≈ 1715 m

How long is the monorail? From the Law of Sines,

b/sin B = c/sin C b = c sin B/sin C

b = 1800 × sin 38°/sin 75°

Monorail: b ≈ 1147 m

How wide is the river?

w = b sin A

w ≈ 1147 sin 67°

River width: w ≈ 1056 m

3 Find the other elements of a triangle with B = 117°, a = 16 cm, and b = 25 cm. This is the SSA (or ASS) case; refer to the table of possibilities for SSA. This problem belongs in row 1, column 2: the opposite side is longer than the adjacent side, and the known angle is > 90°. So there is one and only one solution.

You can use the Law of Sines to find angle A:

(sin A)/a = (sin B)/b

sin A = (a/b) sin B

sin A = (16/25) sin 117° ≈ 0.57024

A ≈ 34.8°

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Find the third angle by subtracting:

C = 180° − A − B

C ≈ 180° − 34.8° − 117°

C ≈ 28.2°

Finally, use the Law of Sines to find side c:

c/sin C = b/sin B

c = b sin C/sin B

c ≈ 25 sin 28.2°/sin 117°

c ≈ 13.3 cm

4 A very modern-looking trivet is a triangular shape with sides of 6″, 9″, and 12″. What are the three angles? This is the SSS case. First, use the Law of Cosines to find angle A:

cos A = (b² + c² − a²) / 2bc

cos A = (9² + 12² − 6²)/(2 × 9 × 12) = 0.875

A ≈ 29.0°

Next, use the Law of Sines to find angle B:

(sin B)/b = (sin A)/a sin B = (b/a) sin A

sin B ≈ (9/6) sin 29.0° ≈ 0.72618

B ≈ 46.6°

Finally, subtract the two angles from 180° to find the third angle:

C = 180° − A − B

C ≈ 180° − 29.0° − 46.6°

C ≈ 104.5°

That’s not a typo, by the way. A and B both happened to round up, but I used the unrounded values to find C. You should never use rounded numbers in further calculations.

5After you’ve painted your bedroom, you have enough paint left to cover 25 ft². You decide to paint a triangle on the wall of another room, as an accent. Two of the angles should be 30° and 40°. Find the third angle, and the lengths of the three sides. The third angle is 180° − 30° − 40°, so C = 110°. Make your sketch using those three angles. (I did this one without measuring the angles, so it’s not perfect. But sketches don’t need to be perfect, just reasonably close.)

Find side c by using equation 32:

c = √2 area × sin C/(sin A sin B)

c = √2 × 25 × sin 110°/(sin 30° sin 40°)

c ≈ 12.1 ft

Then the Law of Sines gets you the other two sides:

a = c sin A/sin C

a ≈ 12.1 sin 30°/sin 110°

a ≈ 6.4 ft

And

b = c sin B/sin C

b ≈ 12.1 sin 40°/sin 110°

b ≈ 8.3 ft

6 You drive 6.0 miles along a straight highway, then take an exit. It’s a right turn, but you don’t notice the angle.

You’re now driving along a straight side road. At the end of 9.8 miles on the side road, you turn 135° to the right, on a third road. (If you’re visualizing this from above, the 135° change of direction corresponds to an angle of 180° − 135° = 45° in the triangle.)

Assuming that road continues in the same direction, how far must you drive to reach your starting point? This is a difficult sketch to draw, because you don’t know the angle of the first turn. But the description gives you two sides and a non-included angle; this is the problematic SSA case. You don’t know exactly where side c will meet side a. To be more precise, you don’t even know if they will meet.

Is it possible for them to meet? Referring back to the table of possibilities within SSA, we see that we’re in the third row, first column: the adjacent side (9.8) is longer than the opposite side (6.0), and the known angle (45°) is < 90°. Compute

h = b sin A

h = 9.8 sin 45° ≈ 6.9 miles

What’s the significance of this? The shortest distance from point C to side c is the segment that meets side c in a right angle. In other words, to get from point C to side c, the shortest possible distance is 6.9 miles. But side a is only 6.0 miles long, so it can never meet side c.

This problem has no solution.

7You’re laying out a triangular bed for your garden. Two sides are 40 m and 60 m, and the angle between them is 22°. How long is the third side, and what are the other two angles? You know two sides and the angle between them. You can use the Law of Cosines to get the third side:

a² = b² + c² − 2bc × cos A

a² = 60² + 40² − 2 × 60 × 40 × cos 22°

c²≈ √749.52c ≈ 27.4 m

Next, for angle B you can use the Law of Sines:

(sin B)/b = (sin A)/a ⇒ sin B = (b/a) sin A

sin B ≈ (60/27.4) sin 22° ≈ 0.82099

Your calculator gives about 55.2° as the angle whose sine is 0.82099, but that looks wrong from the sketch. Clearly B needs to be an obtuse angle, so you remember that sin(180° − x) = sin x, and you subtract 180° − 55.2° to get

B ≈ 124.8°

See how important a sketch is? Of course your sketch probably isn’t perfectly accurate, so you treat it as a device to point out that something might be wrong, but then you look for a way to confirm it. In this case, you have two ways to confirm it:

• You can use the Law of Cosines, which automatically accounts for obtuse angles:

cos B = (a² + c² − b²)/(2ac)

cos B ≈ (27.4² + 40² − 60²)/(2 × 27.4 × 40)

cos B ≈ −0.57095 ⇒ B ≈ 124.8°

• Or, you can compute angle C (below) and then recall that angle C must be < B because side c is < b. If you made C obtuse, 180° − 33.2° = 146.8°, it would be > B whether B is acute or obtuse. So C must be < 90° and B must be > 90°.

Now turn to angle C:

(sin C)/c = (sin A)/a sin C = (c/a) sin A

sin C ≈ (40/27.4) sin 22° ≈ 0.54732

C ≈ 33.2°

## Solutions for Part 5: Functions of Any Angle

1 In which quadrant does the angle −868° occur? What about 42 radians? What are the signs of their sines, cosines, and tangents?

Solution: Your task always is to cast out multiples of ±360° or ±2π, so that you’re left with a positive angle between 0° and 360° (0 and 2π).

−868° = −1080° + 212°. And 212° is between 180° and 270°, so −868° occurs in Q III.

42 radians is about 13.37π, or 12π + 1.37π. 1.37π is obviously between π and 3π/2 (1.5π), so 42 (radians) is in Q III.

In Q III, x and y are both negative. Therefore In Q III, sine and cosine are negative and tangent is positive.

2Rewrite using the smallest possible positive angle of the same trig function:   (a) sin 700°   (b) tan 780°   (c) cos(−390.5) (That’s radians, since there’s no degree mark, but be careful! The angle is −390.5, not −390.5π.)

Solution: (a) 700° = 360° + 340°, so sin 700° = sin 340°. 340° is in Q IV, where y is negative; therefore the sine is negative. The reference angle is 20°, so sin 700° = −sin 20°.

(b) 780° = 720° + 60°, so tan 780° = tan 60°.

(c) −390.5 ≈ −124.3π, or −126π + 1.7π; therefore cos(−390.5) ≈ cos 1.7π. The angle 1.7π radians is between 1.5π and 2π, so it’s in Q IV, where x is positive and therefore the cosine is positive. The reference angle is about 2π − 1.7π ≈ 0.3π or 0.94, so cos(−390.5) ≈ cos 0.94.

3Rewrite as a function of just A:   (a) cos(720° − A)   (b) sin(43π + A)

Solution: (a) 720° is a multiple of 360°, so cos(720° − A) = cos(−A). But cos(−A) = cos A, and therefore cos(720° − A) = cos A.

(b) sin(43π + A) = sin(42π + π + A) = sin(π + A). sin(π + A) = −sin A, so sin(43π + A) = −sin A.

## Solutions for Part 6: The “Squared” Identities

1 If sin A = 3/4, find cos A.

Solution:

sin² A + cos² A = 1

cos² A = 1 − sin² A = 1 − (3/4)² = 1 − 9/16

cos² A = 7/16

cos A = ±√7/16 cos A = √7/4 or −√7/4

Did you remember the ± sign? Just as x² = 9 has two solutions, 3 and −3, so any equation in cos² A has two solutions.

This makes sense in terms of the functions. There are two angles in the interval [0, 2π) or [0°, 360°) where sin A = 3/4, one acute and one obtuse. The acute one has a positive cosine, and the obtuse one has a negative cosine.

2 tan B = −2√2. Find sec B.

Solution: Since this chapter is about the “squared” identities, you can be pretty sure that one exists that connects tan x and sec x. But suppose you met this problem in a different context?

Well, you know two identities involving the tangent function. The definition, tan B = (sin B)/cos B, doesn’t do you a lot of good because it’s got sine and cosine mixed together. You have a dim memory of a squared identity (if you’re like me, it’s dim), but you also know you can re-create it easily, from the one squared identity that you can’t possibly forget:

sin² B + cos² B = 1

You have sin² B, which will become tan² B if you divide both sides by cos² B.

(sin² B)/cos² B + (cos² B)/cos² B = 1/cos² B

tan² B + 1 = sec² B

Success! You have an identity that connects the tangent and secant functions. Now you can proceed to solve the problem.

(−2√2)² + 1 = sec² B

sec² B = (4 × 2) + 1 = 9

sec B = ±3

You need the ± sign because both (x)² and (−x)² are x². But does it make sense in terms of trig? Yes, because the tangent is negative in Q II and Q IV, while the secant—which has the same sign as the cosine, being 1 over the cosine—is positive in Q IV but negative in Q II.

3 tan C = √15. Find cos C.

Solution: Wait, what? You don’t have an identity connecting tangent and cosine. But you do have one connecting tangent and secant, and you know that the secant is 1 over the cosine, so you can do this one too.

tan² C + 1 = sec² C

(√15)² + 1 = 15 + 1 = 16 sec² C = 16

sec C = ±4 cos C = 1/4

4 tan D =√15, Find sin D.

Solution: Okay, in the previous problem you got from tangent to cosine. But you already know how to get from cosine to sine, so you just have one more link in the chain.

tan² D + 1 = sec² D

sec² D = (√15)² + 1 = 16

sec D = 4 cos D = ±1/4

sin² D + cos² D = 1

sin² D = 1 − cos² D = 1 − (±1/4)² =1 − 1/16 =15/16

sin D = ±√15/4

5Prove: sin² x = tan² x / (tan² x + 1)
This assumes that x ≠ π/2 + kπ, for integer k—or 90° + 180k°, if you prefer—because the tangent is undefined for those angles.

Proof:

tan x = tan x

tan x cos x = tan x cos x

sin x = tan x / sec x

sin² x = tan² x / sec² x

But tan² x + 1 = sec² x, so substituting you have

sin² x = tan² x / (tan² x + 1) QED

Many other proofs are possible. For example, you could replace tan² x with sin² x / cos² x in the identity you were asked to prove, and then simplify the fraction. As long as every step in your proof is valid both backwards and forwards, your proof is fine.

If your proof starts with an obvious identity and then works forward to the identity you we asked to prove, as the above proof did, then it only matters that all the steps are valid in that forward direction.

## Solutions for Part 7: Sum and Difference Formulas

1 Find sin(−15°) exactly.

sin(−15°) = sin(30°−45°)

sin(−15°) = sin 30° cos 45° − cos 30° sin 45°

sin(−15°) = (1/2)(√2/2) − (√3/2)(√2/2) = √2/4 − √6/4

sin(−15°) = (√2 − √6)/4

Alternative solution: You could also do this using 45°−60°:

sin(−15°) = sin(45°−60°)

sin(−15°) = sin 45° cos 60° − cos 45° sin 60°

sin(−15°) = (√2/2)(1/2) − (√2/2)(√3/2) = √2/4 − √6/4 = (√2 − √6)/4

2 Find tan 105° exactly.

Solution: 105° = 60° + 45°

tan 105° = tan(60° + 45°)

tan 105° = (tan 60° + tan 45°)/(1 − tan 60° tan 45°)

tan 105° = (√3 + 1)/(1 − (√3)(1)) = (√3 + 1)/(1 − √3)

Multiply top and bottom by 1 + √3 to rationalize the denominator:

tan 105° = (√3 + 1)²/((1 − √3) (1 + √3))

tan 105° = (3 + 2√3 + 1)/(1 − 3) = (4 + 2√3)/(−2) = −2 − √3

3 Prove: cos 2A = 2 cos² A − 1. (Hint: 2A = A + A.)

Proof:

cos 2A = cos(A + A)

cos 2A = cos A cos A − sin A sin A

cos 2A = cos² A − sin² A

But sin² A = 1 − cos² A

cos 2A = cos² A − (1 − cos² A) = cos² A − 1 + cos² A

cos 2A = 2 cos² A − 1 QED

4 Prove these formulas from equation 22, by using the formulas for functions of sum and difference.

(a) cos(−A) = cos A   (I’ve done the first step for you.)

cos(−A) = cos(0 − A)

cos(−A) = cos 0 cos A + sin 0 sin A

cos(−A) = 1 cos A + 0 sin A

cos(−A) = cos A QED

(b) tan(π + A) = tan A

tan(π + A) = (tan π + tan A) / (1 − tan π tan A)

tan(π + A) = (0 + tan A) / (1 − 0 tan A)

tan(π + A) = (tan A) / 1

tan(π + A) = tan A QED

(c) sin(π − A) = sin A

sin(π − A) = sin π cos A − cos π sin A

sin(π − A) = 0 cos A − (−1) sin A

sin(π − A) = sin A QED

## Solutions for Part 8: Double Angle and Half Angle Formulas

1 Use the half-angle formulas to find sin 90° and cos 90°. Of course you already know those; this problem is just for practice in working with the formulas and easy numbers.

Solution: : 90° is half of 180°. The sine and cosine are positive or zero at 90°, so the ± signs in the formulas can be treated as positive.

sin(180°/2) = √(1 − cos 180°)/2 = √(1 − (−1))/2 = √2/2

sin 90° = 1

cos(180°/2) = √(1 + cos 180°)/2 = √(1 + (−1))/2 = √0/2

cos 90° = 0

Why didn’t I ask you to do the same for tan 90°?

2Use the double-angle formulas to find sin 120°, cos 120°, and tan 120° exactly. Again, you already know these; you’re just getting comfortable with the formulas.

Solution: 120° is double 60°. sin 60° = √3/2, cos 60° = 1/2, and tan 60° = √3.

sin(2×60°) = 2 sin 60° cos 60° = 2 (√3/2) (1/2)

sin 120° = (√3)/2

cos(2×60°) = 2 cos² 60° − 1 = 2 × (1/2)² − 1 = (1/2) − 1

cos 120° = −1/2

tan(2×60°) = 2 tan 60° / (1 − tan² 60°) = 2 √3 / (1 − (√3)²) = 2 √3 / (1 − 3) = 2 √3 / (−2)

tan 120° = −√3

Alternative solution: You could also divide:

tan 120° = sin 120° / cos 120° = (√3/2) / (−1/2) = −√3

3 3A = 2A + A. Use the double-angle formulas along with the formulas for sine or cosine of a sum to find formulas for sin 3A in terms of sin A only, and cos 3A in terms of cos A only.

(This is actually done, in a later section, by using a different method.)

Solution: First the sine:

sin(2A + A) = sin 2A cos A + cos 2A sin A

sin 3A = (2 sin A cos A) cos A + (1 − 2 sin² A) sin A

sin 3A = 2 sin A cos² A + sin A − 2 sin³ A

sin 3A = 2 sin A (1 − sin² A) + sin A − 2 sin³ A

sin 3A = 2 sin A − 2 sin³ A + sin A − 2 sin³ A

sin 3A = 3 sin A − 4 sin³ A, or (3 − 4 sin² A) sin A

Now the cosine:

cos(2A + A) = cos 2A cos A − sin 2A sin A

cos 3A = (2 cos² A − 1) cos A − (2 sin A cos A) sin A

cos 3A = 2 cos³ A − cos A − 2 sin² A cos A

cos 3A = 2 cos³ A − cos A − 2 (1 − cos² A) cos A

cos 3A = 2 cos³ A − cos A − 2 cos A  + 2 cos³ A

cos 3A = 4 cos³ A − 3 cos A, or (4 cos² A − 3) cos A

4 Given sin 3A = (3 − 4 sin² A) sin A and cos 3A = (4 cos² A − 3) cos A, find tan 3A in terms of tan A only. Check yourself by computing tan(2A+A).

Solution:

tan 3A = sin 3A / cos 3A

tan 3A = ((3 − 4 sin² A) sin A) / ((4 cos² A − 3) cos A)

tan 3A = (sin A / cos A) (3 − 4 sin² A) / (4 cos² A − 3)

tan 3A = tan A (3 − 4 sin² A) / (4 cos² A − 3)

Divide top and bottom by cos² A. That at least gets rid of sin² A and cos² A, even though it introduces secant functions. (You remember that 1/cos A = sec A, right?)

tan 3A = tan A (3 sec² A − 4 tan² A) / (4 − 3 sec² A)

This may not look better, but it is—before, you had to get rid of two unwanted functions; now you have only one unwanted function, even though it occurs twice. And it’s sec² A. Isn’t there some sort of Pythagorean identity involving sec² A? Yes, there is! sin² A + cos² A = 1 ⇒ tan² A +1 = sec² A.

tan 3A = tan A (3 (tan² A + 1) − 4 tan² A) / (4 − 3 (tan² A + 1))

tan 3A = tan A (3 tan² A + 3 − 4 tan² A) / (4 − 3 tan² A − 3)

tan 3A = tan A (3 − tan² A) / (1 − 3 tan² A)

Check:

tan(2A+A) = (tan 2A + tan A) / (1 − tan 2A tan A)

From equation 60, tan 2A = 2 tan A / (1 − tan² A).

tan 3A = (2 tan A/(1 − tan² A) + tan A) / (1 − (2 tan A/(1 − tan² A)) tan A)

Well, that’s a mess! Clean it up by multiplying top and bottom by (1 − tan² A).

tan 3A = (2 tan A + (1 − tan² A) tan A) / ((1 − tan² A) − 2 tan A tan A)

tan 3A = (3 tan A − tan³ A) / (1 − 3 tan² A)

Factor out tan A from the top of the fraction, and it’s the same as what you got with the first method:

tan 3A = tan A (3 − tan² A) / (1 − 3 tan² A)

5 Find the sine, cosine, and tangent of π/8, exactly.

Solution: π/8 (22½°) is half of π/4 (45°), so you want the half-angle formulas. And π/8 is in Quadrant I, so all function values will be positive.

sin π/8 = √(1 − cos π/4)/2

sin π/8 = √(1 − √2/2)/2 = √(2 − √2)/4 = (½)√2 − √2

This expression contains nested radicals. Though some nested radicals can be denested, following the technique in Denesting Radicals (or Unnesting Radicals), this one unfortunately cannot.

cos π/8 = √(1 + cos π/4)/2

cos π/8 = √(1 + √2/2)/2 = √(2 + √2)/4 = (½)√2 + √2

Now turn to the tangent. You can use the half-angle formula, or simply divide sine by cosine.

tan π/8 = (sin π/8) / (cos π/8)

tan π/8 = √2 − √2 / √2 + √2

This one can be denested, if you multiply top and bottom by √2 − √2. You’d normally do that anyway, to rationalize the denominator, but it’s a nice bonus that that happens to clean up the numerator also.

tan π/8 = √(2 − √2) (2 − √2) / √(2 + √2) (2 − √2)

tan π/8 = (2 − √2) / √(4 − 2) = (2 − √2) / √2 = √2 − 1

## Solutions for Part 9: Inverse Functions

1 The possible output values of Arcsin x include ±π/2, but the possible output values of Arctan x do not. Why can Arctan x never equal −π/2 or π/2?

Solution: Arctan x is an angle whose tangent is x, so the possible values of Arctan x are angles in Q IV and Q I whose tangents can be taken. But tan −π/2 and tan π/2 don’t exist, so they are not possible values of Arctan x.

2 Find sec(Arcsin x). Remember to make a sketch to help you. Pick a value, like x = −0.7, as a test case to check your answer. Solution: If angle A is Arcsin x, then we can make x the opposite side and 1 the hypotenuse. By the theorem of Pythagoras, that makes the adjacent side √1 − x². sec A is therefore sec(Arcsin x) = 1/√1 − x².

There are no odd powers of x in the answer, so we don’t have to worry about the sign of x.

Let’s check that with x = −0.7: Arcsin −0.7 ≈ −44.43°, and sec −44.3° = 1/cos −44.3° ≈ 1.40. 1/√(1 − 0.7²) ≈ 1.40 also, so the formula is right, at least for this test case.

3 Find sin(Arccos 1/x). Remember to make a sketch to help you. Pick a value, like x = 1.3, as a test case to check your answer. Solution: If angle A is Arccos 1/x, we can set the hypotenuse to x and the adjacent side to 1 so that cos A = 1/x as required. The third side is √x² − 1.

What’s sin A? It must be √x² − 1 / x. But that expression has an odd power of x, so we need to check the signs. If x is negative, then Arccos x will be in Q II. The sine function has all positive values in Q II, so we should have a positive answer. But as written, the fraction is negative if x is negative, so it needs an absolute-value sign, just as in Example 3. Our final answer is

sin(Arccos 1/x) = √(x² − 1) / | x |

Let’s check that with x = −1.3: Arccos(1/−1.3) ≈ 140.28°, in Q II as expected, and sin 140.28° ≈ 0.64, a positive number like the sines of all angles in Q II. √((−1.3)² − 1) / | −1.3 | ≈ 0.64 also. So the formula is right, at least for this test case — even with negative x, putting the angle in Q II, the formula returns a positive number, as it needs to.

## Solutions for Part 10: Fun with Complex Numbers

1 Express in a+bi form: (a) 62∠240° (b) 100e1.17i

Solutions: (a) 62(cos(240°) + i sin(240°)) ≈ −31−53.69i

(b) is apparently in radian measure, since there’s no degree mark. 100(cos(1.17) + i sin(1.17)) ≈ 39.02+92.08i

2 Express in polar form, in both degrees and radian measure: (a) −42+17i (b) 100i (c) −14.7

Solutions: (a) r = √(−42)² + 17² ≈ 45.31. θ = 2 Arctan(17/(−42+45.31)) ≈ 2.76; multiply by 180°/π to convert to 157.96°. Answers: 45.31 cis 157.96° or 45.31 cis 2.76. Of course you could use any of the other forms shown in the chapter.

(b) and (c) are “gimmes”, since you don’t have to compute the radius or the angle.
(b) 100i = 100∠90° or 100eiπ/2
(c) −14.7 = 14.7∠180° or 14.7 e

3 Find the three cube roots of −i, in a+bi form. You’ll be able to give an exact answer, not rounded decimals.

Solution: First, put −i into polar form, which is easy enough since −i is on the negative y axis: −i = 1 cis 3π/2. Then apply the formula, equation 83:

(1 cis 3π/2)1/3 = 11/3 cis (π/2 + 2πk/3) for k = 0, 1, 2

The three angles are π/2, π/2 + 2π/3 = 7π/6, and π/2 + 4π/3 = 11π/6 = 3π/2.

(1 cis 3π/2)1/3 = 1 cis π/2, 1 cis 7π/6, 1 cis 11π/6.

You know exact sines and cosines for any multiple of π/6, so you don’t need your calculator to convert back to a+bi form:

Cube roots of −i: i, −(√3/2)−(1/2)i, (√3/2)−(1/2)i.

Incidentally, not only is i a cube root of −i, but −i is a cube root of i. I’ll let you find the other two cube roots of i yourself.

4 Find (1.04−0.10i)16.

Solution: First, put that number into polar form: r = √1.04²+0.10² ≈ 1.044796631. Then, find θ = 2 Arctan(−0.10/(1.04+1.044796631)) ≈ −0.0958591471. (My calculator happens to be in radian mode, but it doesn’t matter because I’ll be converting back to rectangular format anyway.) So

(1.04−0.10i)≈ 1.04796631 cis −0.0958591471

And therefore

(1.04−0.10i)16≈ 1.0479663116 cis (−16;0.0958591471)

≈ 2.016082111 cis (−1.533746354)

≈ 0.0746787−2.01469853i

## What’s New

• 6 Nov 2020: Converted page from HTML 4.01 to HTML5, and improved the formatting of radicals.
• 20 Dec 2016: In a problem that produces nested radicals, added a reference to a technique on this site for denesting them.
• 11 Dec 2016: Added solutions for practice problems for Part 5; added three new proofs in Part 7.
• 29 Nov 2016: Added three new problems in Part 3.
• 26/27 Nov 2016: Added solutions for Part 7 and Part 8 problems.
• 19/20 Nov 2017: Added solutions for Part 1, Part 3, and Part 6 problems.
• 13 Nov 2016: Added solutions for Part 5 problems. (Part 5 was renumbered as Part 4 on 29 Nov 2016.)
• 30 Oct 2016: Added solutions for Part 9 and Part 10 problems.
• 27 Oct 2016: New document (Part 2 problems only).
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