Trig without Tears:
Solutions to Practice Problems
Copyright © 1997–2023 by Stan Brown, BrownMath.com
Trig without Tears:
Copyright © 1997–2023 by Stan Brown, BrownMath.com
Solutions:
(a) (π/6) × (180°/π) = 30°
(b) 2π × (180°/π) = 360°
(c) 1 × (180°/π) = (180/π)° ≈ 57.3°
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°).
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°.
Solution: Cue the Pythagorean Theorem!
c² = a² + b²
c² = 5² + 12²
c² = 25 + 144 = 169
c = √169 = 13
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.
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?)
Answers: Quadrant I: 12 and 3; Quadrant II: 9 and 12; Quadrant III: 6 and 9; Quadrant IV: 3 and 6.
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°.
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°.
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.
(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.
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.
Answers: sin(π/4) = √2/2 or 1/√2; cos(π/6) = √3/2; tan(π/3) = √3
Answers:
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 = √3 ⇒ H = 60° or π/3
cos I = 1 ⇒ I = 0° or 0 [radians]
cos J = 0 ⇒ J = 90° or π/2
Solutions: sec 60° = 1/(cos 60°) = 1/(1/2) ⇒ sec 60° = 2
cot 30° = 1/(tan 30°) = 1/(√3/3) = 3/√3 = √3 ⇒ cot 30° = √3
You could also do the second one using equation 6:
cot 30° = tan(90°−30°) = tan 60° = √3
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
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
Solution: tan 150° = −tan(180° − 150°) = −tan 30° = −√3/3
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° − A ⇒ B ≈ 67.2°
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
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°
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
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.
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
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.
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≈ √749.52 ⇒ c ≈ 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:
cos B = (a² + c² − b²)/(2ac)
cos B ≈ (27.4² + 40² − 60²)/(2 × 27.4 × 40)
cos B ≈ −0.57095 ⇒ B ≈ 124.8°
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°
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.
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.
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.
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.
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.
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.
Start with the squared identity from the previous problem:
tan² C + 1 = sec² C
(√15)² + 1 = 15 + 1 = 16 ⇒ sec² C = 16
sec C = ±4 ⇒ cos C = 1/4
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
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.
Solution: Start with −15° = 30° − 45°
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
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
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
(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
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°?
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
(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
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)
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
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.
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.
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: (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
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 eiπ
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.
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
Updates and new info: https://BrownMath.com/twt/