IEB Trigonometry practice

(a) Answer: sin A = 5/13

1. A lies in the second quadrant: x = −12, r = 13
2. y^{^2} = 13^{^2} − (−12)^{^2} = 25, so y = 5
3. sin A = 5/13 and cos A = −12/13

(b) Answer: sin(A + B) = 120/169

1. B lies in the fourth quadrant: x = 12, r = 13
2. y^{^2} = 13^{^2} − 12^{^2} = 25, so y = −5
3. sin B = −5/13 and cos B = 12/13
4. sin(A + B) = sin A cos B + cos A sin B
5. = (5/13)(12/13) + (−12/13)(−5/13)
6. = 120/169

(c) Answer: sin 2A = −120/169

1. sin 2A = 2 sin A cos A
2. = 2(5/13)(−12/13)
3. = −120/169

(a) Answer: −cos θ

1. sin(360° − θ) = −sin θ
2. sin(90° − θ) = cos θ
3. sin(180° − θ) = sin θ
4. = −cos θ

(b) Answer: −√3/2

1. cos 89° cos 61° − sin 89° sin 61° = cos(89° + 61°)
2. = cos 150° = −√3/2

(c) Answer: 1/2

1. 2 sin 15° cos 15° = sin 30°
2. = 1/2

(d) Answer: θ = −30° + k·180°, k ∈ ℤ

1. tan θ = −1/√3

(a) Answer: x = −120° + k·360° or x = 0° + k·360° or x = 120° + k·360°, k ∈ ℤ

1. 2sin^{^2} x + cos x − 1 = 0
2. 2(1 − cos^{^2} x) + 1 cos x − 1 = 0
3. 2cos^{^2} x − cos x − 1 = 0
4. (2cos x + 1)(cos x − 1) = 0
5. cos x = −1/2 or cos x = 1

(b) Answer: See the proof below.

1. LHS = (2cos^{^2} x − 1 + 1)/(cos x)
2. = (2cos^{^2} x)/(cos x)
3. = 2 cos x
4. = RHS

(c) Answer: no solution

1. 2 cos x = 0, so cos x = 0
2. x = ±90° + k·360°
3. x = −90° or x = 90°: the identity is not valid there
4. In [−180°; 180°]: no solution

(a) Answer: See the proof below.

1. LHS = (cos^{^2} x − sin^{^2} x + sin^{^2} x)/(cos x)
2. = (cos^{^2} x)/(cos x)
3. = cos x
4. = RHS

(b) Answer: x = 90° or x = 270°

1. The identity is not valid where cos x = 0.
2. cos x = 0: x = 90° + k·180°
3. In [0°; 360°]: x = 90° or x = 270°

(c) Answer: no solution

1. cos x = 0
2. x = ±90° + k·360°
3. x = −90° or x = 90°: the identity is not valid there, so these are not solutions
4. In [−180°; 180°]: no solution

(a) Answer: sin 314° = −p

1. sin 314° = sin(360° − 46°) = −sin 46° = −p

(b) Answer: cos 46° = √(1 − p^{^2})

1. cos^{^2} 46° = 1 − sin^{^2} 46° = 1 − p^{^2}
2. 46° is acute, so cos 46° > 0

(c) Answer: sin 92° = 2p√(1 − p^{^2})

1. sin 92° = 2 sin 46° cos 46°
2. = 2p√(1 − p^{^2})

(d) Answer: sin 83° cos 37° − cos 83° sin 37° = p

1. sin 83° cos 37° − cos 83° sin 37° = sin(83° − 37°)
2. = sin 46° = p

(a) Answer: sin 72° = 2p

1. sin 72° = 2 sin 36° cos 36° = 2p

(b) Answer: cos 18° = 2p

1. cos 18° = cos(90° − 72°) = sin 72° = 2p

(c) Answer: cos 144° = 1 − 8p^{^2}

1. cos 144° = 1 − 2sin^{^2} 72°
2. = 1 − 2(2p)^{^2} = 1 − 8p^{^2}

(d) Answer: cos 72° = √(1 − 4p^{^2})

1. cos^{^2} 72° = 1 − sin^{^2} 72° = 1 − 4p^{^2}
2. 72° is acute, so cos 72° > 0

(a) Answer: a = 3 and p = −45°

1. the greatest value of f is 3, so a = 3
2. cos(x + p) is greatest where x + p = 0°
3. 45° + p = 0°, so p = −45°

(b) Answer: −3 ≤ y ≤ 3

1. f runs from −3 to 3

(c) Answer: See the sketch below.

1. g(x) = 3 sin x
2. g: x-intercepts (−180°; 0), (0°; 0), (180°; 0)
3. g: y-intercept (0°; 0)
4. g: turning points (−90°; −3), (90°; 3)
5. g: end points (−180°; 0), (180°; 0)

(d) Answer: x = −112.5° or x = 67.5°

1. 3 cos(x − 45°) = 3 sin x
2. cos(x − 45°) = cos(90° − x)
3. x − 45° = 90° − x + k·360°, so x = 67.5° + k·180°
4. or x − 45° = −(90° − x) + k·360°, which gives −45° = −90° + k·360°: no solution
5. In [−180°; 180°]: x = −112.5° or x = 67.5°

(e) Answer: −112.5° < x < 67.5°

1. f lies above g, read from the sketch

(a) Answer: x = 120° + k·360° or x = −120° + k·360° or x = k·360°

1. cos 2x = cos x
2. 2cos^{^2} x − 1 = cos x
3. 2cos^{^2} x − cos x − 1 = 0
4. (2 cos x + 1)(cos x − 1) = 0
5. cos x = −1/2: x = ±120° + k·360°
6. or cos x = 1: x = k·360°

(b) Answer: x = −120° or x = 0°

1. k = …, −1, 0, 1, … in x = 120° + k·360° or x = −120° + k·360° or x = k·360°
2. In [−180°; 90°]: x = −120° or x = 0°

(c) Answer: See the sketch below.

1. f(x) = cos 2x
2. f: x-intercepts (−135°; 0), (−45°; 0), (45°; 0)
3. f: y-intercept (0°; 1)
4. f: turning points (−90°; −1), (0°; 1)
5. f: end points (−180°; 1), (90°; −1)
6. g(x) = cos x
7. g: x-intercepts (−90°; 0), (90°; 0)
8. g: y-intercept (0°; 1)
9. g: turning points (0°; 1)
10. g: end points (−180°; −1), (90°; 0)

(d) Answer: −180° ≤ x < −135° or −90° < x < −45° or 45° < x < 90°

1. f and g have opposite signs, and g(x) ≠ 0, read from the sketch

(a) Answer: x = 90° and x = 270°

1. tan x is undefined where cos x = 0: x = 90° + k·180°
2. In [0°; 360°]: x = 90° and x = 270°

(b) Answer: x = 0° or x = 45° or x = 135° or x = 180° or x = 225° or x = 315° or x = 360°

1. sin x/cos x = 2 sin x cos x
2. sin x = 2 sin x cos^{^2} x (cos x ≠ 0)
3. sin x(1 − 2cos^{^2} x) = 0
4. sin x = 0: x = k·180°
5. or cos^{^2} x = 1/2, cos x = ±√2/2: x = 45° + k·90°
6. In [0°; 360°]: x = 0° or x = 45° or x = 135° or x = 180° or x = 225° or x = 315° or x = 360°

(c) Answer: See the sketch below.

1. f(x) = sin 2x
2. f: x-intercepts (0°; 0), (90°; 0), (180°; 0), (270°; 0), (360°; 0)
3. f: y-intercept (0°; 0)
4. f: turning points (45°; 1), (135°; −1), (225°; 1), (315°; −1)
5. f: end points (0°; 0), (360°; 0)
6. g(x) = tan x
7. g: x-intercepts (0°; 0), (180°; 0), (360°; 0)
8. g: y-intercept (0°; 0)
9. g: asymptotes x = 90°, x = 270°
10. g: end points (0°; 0), (360°; 0)

(d) Answer: 0° ≤ x < 90° or 90° < x < 270° or 270° < x ≤ 360°

1. f and g have the same sign, or one of them is 0, read from the sketch

(a) Answer: 180°

1. the period of cos bx is 360°/b = 360°/2 = 180°

(b) Answer: −1 ≤ y ≤ 1

1. f runs from −1 to 1

(c) Answer: g(x) = cos(2x − 120°)

1. g(x) = f(x − 60°) = cos 2(x − 60°)

(d) Answer: See the sketch below.

1. f(x) = cos 2x
2. f: x-intercepts (−45°; 0), (45°; 0)
3. f: y-intercept (0°; 1)
4. f: turning points (0°; 1)
5. f: end points (−90°; −1), (90°; −1)
6. g(x) = cos(2x − 120°)
7. g: x-intercepts (−75°; 0), (15°; 0)
8. g: y-intercept (0°; −1/2)
9. g: turning points (−30°; −1), (60°; 1)
10. g: end points (−90°; 1/2), (90°; 1/2)

(e) Answer: x = −60° or x = 30°

1. cos 2x = cos(2x − 120°)
2. 2x = 2x − 120° + k·360°: no solution, as 120° is not a multiple of 360°
3. or 2x = −(2x − 120°) + k·360°
4. 4x = 120° + k·360°
5. x = 30° + k·90°
6. In [−90°; 90°]: x = −60° or x = 30°

(a) Answer: See the working below.

1. AT makes an angle α and the road an angle θ with the horizontal: P$\hat{\mathrm{A}}$T = α − θ
2. the horizontal at P makes 90° with PT, and PA lies θ below it: A$\hat{\mathrm{P}}$T = 90° + θ
3. A$\hat{\mathrm{T}}$P = 180° − (α − θ) − (90° + θ) = 90° − α (sum ∠s of Δ)

(b) Answer: See the working below.

1. In ΔAPT: PT/(sin P$\hat{\mathrm{A}}$T) = AP/(sin A$\hat{\mathrm{T}}$P) (sine rule)
2. PT = (d sin(α − θ))/(sin(90° − α))
3. sin(90° − α) = cos α, so PT = (d sin(α − θ))/(cos α)

(c) Answer: PT = 8.37 m

1. PT = (28 sin 15°)/(cos 30°)
2. = 8.37 m

(d) Answer: BT = 50.81 m

1. AT = (d sin(90° + θ))/(sin(90° − α)) = (28 cos 15°)/(cos 30°) = 31.23 m (sine rule)
2. B$\hat{\mathrm{A}}$T = 180° − 15° = 165° (∠s on a str line)
3. BT^{^2} = 20^{^2} + 31.23^{^2} − 2(20)(31.23) cos 165° (cosine rule)

(e) Answer: 24.15°

1. (sin A$\hat{\mathrm{B}}$T)/AT = (sin B$\hat{\mathrm{A}}$T)/BT (sine rule)
2. A$\hat{\mathrm{B}}$T = 9.15°
3. the road rises at 15°: the angle of elevation is 9.15° + 15° = 24.15°

(a) Answer: BD = 7.37 cm

1. BD^{^2} = AB^{^2} + AD^{^2} − 2(AB)(AD) cos B$\hat{\mathrm{A}}$D (cosine rule)
2. = 9^{^2} + 5^{^2} − 2(9)(5) cos 55°

(b) Answer: B$\hat{\mathrm{C}}$D = 125° (opp ∠s of cyclic quad)

1. B$\hat{\mathrm{C}}$D = 180° − 55° = 125° (opp ∠s of cyclic quad)

(c) Answer: B$\hat{\mathrm{D}}$C = 19.47°

1. (sin B$\hat{\mathrm{D}}$C)/BC = (sin B$\hat{\mathrm{C}}$D)/BD (sine rule)
2. sin B$\hat{\mathrm{D}}$C = (3 sin 125°)/7.37

(d) Answer: CD = 5.23 cm

1. C$\hat{\mathrm{B}}$D = 180° − 125° − 19.47° = 35.53° (sum ∠s of Δ)
2. CD/(sin C$\hat{\mathrm{B}}$D) = BD/(sin B$\hat{\mathrm{C}}$D) (sine rule)
3. CD = (7.37 sin 35.53°)/(sin 125°) = 5.23 cm

(e) Answer: area ABCD = 24.86 cm²

1. area ΔABD = (1/2)(9)(5) sin 55° (area rule)
2. area ΔBCD = (1/2)(3)(7.37) sin 35.53° (area rule)

(a) Answer: See the working below.

1. In ΔDPA, the right angle at P: tan 30° = h/AP, so AP = h/(tan 30°)
2. In ΔDPB: tan 25° = h/BP, so BP = h/(tan 25°)

(b) Answer: h = 84.26 m

1. AB^{^2} = AP^{^2} + BP^{^2} − 2(AP)(BP) cos A$\hat{\mathrm{P}}$B (cosine rule)
2. 260^{^2} = h^{^2}/(tan^{^2} 30°) + h^{^2}/(tan^{^2} 25°) − 2(h/(tan 30°))(h/(tan 25°)) cos 105°
3. 260^{^2} = 9.5216h^{^2}

(c) Answer: AP = 145.94 m

1. AP = 84.26/(tan 30°) = 145.94 m

(d) Answer: B$\hat{\mathrm{A}}$P = 42.17°

1. BP = 84.26/(tan 25°) = 180.69 m
2. cos B$\hat{\mathrm{A}}$P = (AP^{^2} + AB^{^2} − BP^{^2})/(2(AP)(AB)) (cosine rule)
3. = (145.94^{^2} + 260^{^2} − 180.69^{^2})/(2(145.94)(260))

(e) Answer: area = 12 736.13 m²

1. area = (1/2)(AP)(BP) sin A$\hat{\mathrm{P}}$B
2. = (1/2)(145.94)(180.69) sin 105° = 12 736.13 m²

(a) Answer: AE = 1.88 m

1. E is above the midpoint M of AD, so AM = 1.25 m
2. AE = √(1.25^{^2} + 1.4^{^2}) = 1.88 m

(b) Answer: 48.24°

1. in the end ADE the side meets the floor at an angle θ, with tan θ = 1.4/1.25
2. θ = 48.24°

(c) Answer: AF = 4.42 m

1. F is above the midpoint N of BC: AN = √(4^{^2} + 1.25^{^2}) = 4.19 m
2. AF = √(AN^{^2} + 1.4^{^2}) = 4.42 m

(d) Answer: 18.47°

1. FN is vertical: tan F$\hat{\mathrm{A}}$N = 1.4/AN = 1.4/4.19
2. F$\hat{\mathrm{A}}$N = 18.47°

(e) Answer: A$\hat{\mathrm{F}}$C = 87.25°

1. FC = AE = 1.88 m and AC = √(4^{^2} + 2.5^{^2}) = 4.72 m
2. cos A$\hat{\mathrm{F}}$C = (AF^{^2} + FC^{^2} − AC^{^2})/(2(AF)(FC)) (cosine rule)
3. = (4.42^{^2} + 1.88^{^2} − 4.72^{^2})/(2(4.42)(1.88))

(a) Answer: LA = 163.54 m and LB = 257.23 m

1. the angle of elevation of T from A equals the angle of depression of A from T: L$\hat{\mathrm{A}}$T = 17° (alt ∠s; horizontals ∥)
2. tan 17° = 50/LA, so LA = 50/(tan 17°) = 163.54 m
3. likewise LB = 50/(tan 11°) = 257.23 m

(b) Answer: AB = 402.28 m

1. AB^{^2} = LA^{^2} + LB^{^2} − 2(LA)(LB) cos A$\hat{\mathrm{L}}$B (cosine rule)
2. = 163.54^{^2} + 257.23^{^2} − 2(163.54)(257.23) cos 145°

(c) Answer: B$\hat{\mathrm{A}}$L = 21.52°

1. cos B$\hat{\mathrm{A}}$L = (LA^{^2} + AB^{^2} − LB^{^2})/(2(LA)(AB)) (cosine rule)
2. = (163.54^{^2} + 402.28^{^2} − 257.23^{^2})/(2(163.54)(402.28))

(d) Answer: 33.29 m

1. its distance from L is then 50/(tan 21°) = 130.254 m
2. it sails 163.543 − 130.254 = 33.289 m