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
2. θ = −30° + k·180°, k ∈ ℤ

(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
6. x = −120° + k·360° or x = 0° + k·360° or x = 120° + k·360°, k ∈ ℤ

(b) Answer: See the proof below.

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

(c) Answer: x = 60° or x = 240°

1. 2 tan x = 2√3, so tan x = √3
2. x = 60° + k·180°
3. In [0°; 360°]: x = 60° or x = 240°

(a) Answer: See the proof below.

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

(b) Answer: x = 0° or x = 90° or x = 180° or x = 270° or x = 360°

1. The identity is not valid where sin 2x = 0, or where tan x is undefined.
2. sin 2x = 0: 2x = k·180°, so x = k·90°
3. tan x is undefined where cos x = 0, which is among these values
4. In [0°; 360°]: x = 0° or x = 90° or x = 180° or x = 270° or x = 360°

(c) Answer: no solution

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

(a) Answer: sin 44° = 2p

1. sin 44° = 2 sin 22° cos 22° = 2p

(b) Answer: cos 46° = 2p

1. cos 46° = cos(90° − 44°) = sin 44° = 2p

(c) Answer: cos 88° = 1 − 8p^{^2}

1. cos 88° = 1 − 2sin^{^2} 44°
2. = 1 − 2(2p)^{^2} = 1 − 8p^{^2}

(d) Answer: cos 44° = √(1 − 4p^{^2})

1. cos^{^2} 44° = 1 − sin^{^2} 44° = 1 − 4p^{^2}
2. 44° is acute, so cos 44° > 0
3. cos 44° = √(1 − 4p^{^2})

(a) Answer: sin 134° = p

1. sin 134° = sin(180° − 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
3. cos 46° = √(1 − p^{^2})

(c) Answer: sin 92° = 2p√(1 − p^{^2})

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

(a) Answer: 180°

1. g(x) = −tan x repeats every 180°

(b) Answer: x = −90° and x = 90°

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

(c) Answer: See the sketch below.

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

(d) Answer: x = −180° or −135° ≤ x < −90° or −45° ≤ x ≤ 0° or 45° ≤ x < 90° or 135° ≤ x ≤ 180°

1. f and g have opposite signs, or one of them is 0, read from the sketch
2. x = −180° or −135° ≤ x < −90° or −45° ≤ x ≤ 0° or 45° ≤ x < 90° or 135° ≤ x ≤ 180°

(a) Answer: b = 2

1. sin bx has period 360°/b
2. one full wave of f takes 180°
3. 360°/b = 180°, so b = 2

(b) Answer: 180°

1. f(x) = sin 2x repeats every 180°

(c) Answer: See the sketch below.

1. g(x) = −cos x
2. g: x-intercepts (−90°; 0), (90°; 0), (270°; 0)
3. g: y-intercept (0°; −1)
4. g: turning points (0°; −1), (180°; 1)
5. g: end points (−90°; 0), (270°; 0)

(d) Answer: 1

1. the amplitude of a·cos x is |a| = 1

(e) Answer: x = −90° or x = −30° or x = 90° or x = 210° or x = 270°

1. sin 2x = −cos x
2. 2 sin x cos x + cos x = 0
3. cos x(2 sin x + 1) = 0
4. cos x = 0 or sin x = −1/2
5. x = 90° + k·180° or x = −30° + k·360° or x = 210° + k·360°
6. In [−90°; 270°]: x = −90° or x = −30° or x = 90° or x = 210° or x = 270°

(f) Answer: −90° < x < −30° or 90° < x < 210°

1. f lies below g, read from the sketch
2. −90° < x < −30° or 90° < x < 210°

(g) Answer: 0° < x < 90° or 90° < x < 180°

1. f and g have opposite signs, and g(x) ≠ 0, read from the sketch
2. 0° < x < 90° or 90° < x < 180°

(a) Answer: See the working below.

1. A$\hat{\mathrm{C}}$B = 180° − (α + β) (sum ∠s of Δ)
2. AC/(sin β) = d/(sin(180° − (α + β))) (sine rule)
3. sin(180° − (α + β)) = sin(α + β)
4. AC = (d sin β)/(sin(α + β))

(b) Answer: See the working below.

1. In ΔACT, A$\hat{\mathrm{C}}$T = 90°: tan γ = TC/AC
2. TC = AC tan γ = (d sin β tan γ)/(sin(α + β))

(c) Answer: TC = 96.57 m

1. TC = (100 sin 65° tan 44°)/(sin 115°)
2. = 96.57 m

(d) Answer: BT = 128.33 m

1. BC = (100 sin 50°)/(sin 115°) = 84.52 m (sine rule)
2. BT^{^2} = BC^{^2} + TC^{^2} = 84.52^{^2} + 96.57^{^2}
3. BT = 128.33 m

(e) Answer: 48.81°

1. tan T$\hat{\mathrm{B}}$C = TC/BC = 96.57/84.52
2. T$\hat{\mathrm{B}}$C = 48.81°

(a) Answer: QS = 78.02 m

1. In ΔPQS: (QS)/(sin Q$\hat{\mathrm{P}}$S) = (PQ)/(sin Q$\hat{\mathrm{S}}$P)
2. QS = (100 sin 45°)/(sin 65°)
3. = 78.02 m

(b) Answer: QR = 94.69 m

1. Q$\hat{\mathrm{S}}$R = 180° − 65° = 115° (∠s on a str line)
2. QR^{^2} = QS^{^2} + SR^{^2} − 2(QS)(SR) cos Q$\hat{\mathrm{S}}$R
3. = 78.02^{^2} + 30^{^2} − 2(78.02)(30) cos 115°
4. QR = 94.69 m

(c) Answer: area = 1060.66 m²

1. Area = (1/2)(QS)(SR) sin Q$\hat{\mathrm{S}}$R
2. = (1/2)(78.02)(30) sin 115°
3. = 1060.66 m²

(d) Answer: TQ = 36.38 m

1. In ΔTQS, T$\hat{\mathrm{Q}}$S = 90°: tan 25° = (TQ)/(QS)
2. TQ = 78.02 tan 25°
3. = 36.38 m