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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 11 · 11.8 Unit Circle

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.π3\dfrac{\pi}{3}
2.150∘150^\circ
3.12\dfrac{1}{2}
4.12\dfrac{1}{2}
5.1
6.12 cm
7.36 cm2^2
8.12.6 cm
9.25.1 cm2^2
10.Both equal 22\dfrac{\sqrt{2}}{2}.
Silver
11.−32-\dfrac{\sqrt{3}}{2}
12.−12-\dfrac{1}{2}
13.x=2π3x = \dfrac{2\pi}{3} or 4π3\dfrac{4\pi}{3}
14.45\dfrac{4}{5}
15.34\dfrac{3}{4}
16.30.6 cm
17.32\dfrac{\sqrt{3}}{2}
18.−sin⁡θ-\sin\theta and cos⁡θ\cos\theta.
19.x=2π3x = \dfrac{2\pi}{3} or 5π3\dfrac{5\pi}{3}
20.(a) 4π4\pi cm; (b) 18π18\pi cm2^2
Gold
21.5π4\dfrac{5\pi}{4}
22.−1+32-\dfrac{1 + \sqrt{3}}{2}
23.x=π4,3π4,5π4,7π4x = \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}
24.≈61.4\approx 61.4 cm2^2
25.LHS =sin⁡2θsin⁡θ=sin⁡θ= \dfrac{\sin^2\theta}{\sin\theta} = \sin\theta.
26.cos⁡θ=−35\cos\theta = -\dfrac{3}{5}, tan⁡θ=43\tan\theta = \dfrac{4}{3}
27.x=π12,5π12,13π12,17π12x = \dfrac{\pi}{12}, \dfrac{5\pi}{12}, \dfrac{13\pi}{12}, \dfrac{17\pi}{12}
28.6−24\dfrac{\sqrt{6} - \sqrt{2}}{4}
29.105∘105^\circ
30.x=π6,5π6,3π2x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}
Platinum
31.r=12r = 12; segment ≈13.7\approx 13.7 cm2^2
32.θ=32\theta = \dfrac{3}{2} rad ≈85.9∘\approx 85.9^\circ
33.See working.
34.x=π6,5π6,3π2x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}
35.r=5r = 5 cm; max area 25 cm2^2
36.6+24\dfrac{\sqrt{6} + \sqrt{2}}{4}
37.θ=π3\theta = \dfrac{\pi}{3}; segment =6π−93≈3.26= 6\pi - 9\sqrt{3} \approx 3.26 cm2^2
38.x=0,π2,2πx = 0, \dfrac{\pi}{2}, 2\pi
39.11π2\dfrac{11\pi}{2} cm2^2
40.See working.

Pack B — Answers

Bronze
1.3π4\dfrac{3\pi}{4}
2.120∘120^\circ
3.22\dfrac{\sqrt{2}}{2}
4.32\dfrac{\sqrt{3}}{2}
5.3\sqrt{3}
6.12 cm
7.30 cm2^2
8.6.28 cm
9.52.4 cm2^2
10.sin⁡60∘=32\sin 60^\circ = \dfrac{\sqrt{3}}{2}; cos⁡60∘=12\cos 60^\circ = \dfrac{1}{2}.
Silver
11.−12-\dfrac{1}{2}
12.−22-\dfrac{\sqrt{2}}{2}
13.x=π4x = \dfrac{\pi}{4} or 3π4\dfrac{3\pi}{4}
14.1213\dfrac{12}{13}
15.512\dfrac{5}{12}
16.21.8 cm
17.32\dfrac{\sqrt{3}}{2}
18.−tan⁡θ-\tan\theta; since tan⁡(−θ)=sin⁡(−θ)cos⁡(−θ)=−sin⁡θcos⁡θ\tan(-\theta) = \frac{\sin(-\theta)}{\cos(-\theta)} = \frac{-\sin\theta}{\cos\theta}.
19.x=π4x = \dfrac{\pi}{4} or 5π4\dfrac{5\pi}{4}
20.(a) 10π10\pi cm; (b) 60π60\pi cm2^2
Gold
21.5π3\dfrac{5\pi}{3}
22.−1+32-\dfrac{1 + \sqrt{3}}{2}
23.x=π3,2π3,4π3,5π3x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}
24.≈52.8\approx 52.8 cm2^2
25.LHS =(1−cos⁡2θ)−cos⁡2θ=1−2cos⁡2θ= (1 - \cos^2\theta) - \cos^2\theta = 1 - 2\cos^2\theta = RHS.
26.sin⁡θ=1213\sin\theta = \dfrac{12}{13}, tan⁡θ=−125\tan\theta = -\dfrac{12}{5}
27.x=π8,7π8,9π8,15π8x = \dfrac{\pi}{8}, \dfrac{7\pi}{8}, \dfrac{9\pi}{8}, \dfrac{15\pi}{8}
28.6+24\dfrac{\sqrt{6} + \sqrt{2}}{4}
29.110∘110^\circ
30.x=0,2π3,4π3x = 0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}
Platinum
31.r≈13.07r \approx 13.07; segment ≈4.85\approx 4.85 cm2^2
32.θ=2\theta = 2 rad ≈114.6∘\approx 114.6^\circ
33.See working.
34.x=π6,5π6,π2,3π2x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{\pi}{2}, \dfrac{3\pi}{2}
35.r=7.5r = 7.5 cm; max area 56.25 cm2^2
36.2−64\dfrac{\sqrt{2} - \sqrt{6}}{4}
37.θ=π3\theta = \dfrac{\pi}{3}; segment =32π3−163≈5.79= \dfrac{32\pi}{3} - 16\sqrt{3} \approx 5.79 cm2^2
38.x=π4,5π4x = \dfrac{\pi}{4}, \dfrac{5\pi}{4}
39.2π2\pi cm2^2
40.See working: sin⁡θcos⁡θ=0\sin\theta \cos\theta = 0.

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) 32\frac{\sqrt{3}}{2}. (b) 32\frac{\sqrt{3}}{2}. (c) 1. (d) 12\frac{1}{2}.
Full working
From the 30-60-90 and 45-45-90 special triangles, with π6=30∘\frac{\pi}{6} = 30^\circ.
2Problem 2
Answer
−32-\dfrac{\sqrt{3}}{2}.
Full working
5π6\frac{5\pi}{6} is in QII (between π2\frac{\pi}{2} and π\pi). Reference angle: π−5π6=π6\pi - \frac{5\pi}{6} = \frac{\pi}{6}. Cosine is negative in QII: cos⁡5π6=−cos⁡π6=−32\cos\frac{5\pi}{6} = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}.
3Problem 3
Answer
(a) 4π≈12.64\pi \approx 12.6 cm. (b) 18π≈56.518\pi \approx 56.5 cm2^2. (c) 18+4π≈30.618 + 4\pi \approx 30.6 cm.
Full working
(a) s=rθ=4πs = r\theta = 4\pi. (b) A=12r2θ=18πA = \frac{1}{2}r^2\theta = 18\pi. (c) Perimeter =2r+s=18+4π= 2r + s = 18 + 4\pi.
4Problem 4
Answer
(a) x=2π3x = \frac{2\pi}{3} or 4π3\frac{4\pi}{3}. (b) x=45∘x = 45^\circ or 135∘135^\circ. (c) x=2π3x = \frac{2\pi}{3} or 5π3\frac{5\pi}{3}.
Full working
(a) cos⁡x=−12\cos x = -\frac{1}{2}, ref π3\frac{\pi}{3}, cos negative in QII, QIII. (b) Sin positive in QI, QII. (c) Tan negative in QII, QIV; ref π3\frac{\pi}{3}.
5Problem 5
Answer
(a) −45-\frac{4}{5}. (b) −34-\frac{3}{4}. (c) −1225-\frac{12}{25}.
Full working
(a) cos⁡2θ=1−925=1625\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}. QII → cosine negative: −45-\frac{4}{5}. (b) tan⁡=sin⁡/cos⁡=−34\tan = \sin/\cos = -\frac{3}{4}. (c) Multiply.
6Problem 6
Answer
(a) Major sector 200π3≈209.4\frac{200\pi}{3} \approx 209.4 cm2^2. (b) 253≈43.325\sqrt{3} \approx 43.3 cm2^2. (c) Minor segment ≈61.4\approx 61.4 cm2^2.
Full working
(a) Minor sector =100π3= \frac{100\pi}{3}; major sector =πr2−minor=200π3= \pi r^2 - \text{minor} = \frac{200\pi}{3}. (b) Triangle =12r2sin⁡θ=253= \frac{1}{2}r^2\sin\theta = 25\sqrt{3}. (c) Minor segment =100π3−253≈61.4= \frac{100\pi}{3} - 25\sqrt{3} \approx 61.4.
7Problem 7
Answer
See working.
Full working
Cross-multiplying: (1−cos⁡θ)(1+cos⁡θ)=sin⁡2θ(1 - \cos\theta)(1 + \cos\theta) = \sin^2\theta. LHS =1−cos⁡2θ=sin⁡2θ= 1 - \cos^2\theta = \sin^2\theta by Pythagorean identity. Hence the identity holds, provided sin⁡θ≠0\sin\theta \neq 0 (LHS denominator) and 1+cos⁡θ≠01 + \cos\theta \neq 0 (RHS denominator), i.e. θ≠nπ\theta \neq n\pi.
8Problem 8
Answer
x=π6,5π6,3π2x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}.
Full working
Let u=sin⁡xu = \sin x: 2u2+u−1=(2u−1)(u+1)=02u^2 + u - 1 = (2u - 1)(u + 1) = 0. u=12u = \frac{1}{2} → x=π6,5π6x = \frac{\pi}{6}, \frac{5\pi}{6}. u=−1u = -1 → x=3π2x = \frac{3\pi}{2}.
9Problem 9
Answer
(a) 6+24\frac{\sqrt{6} + \sqrt{2}}{4}. (b) 2−64\frac{\sqrt{2} - \sqrt{6}}{4}.
Full working
sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B with A=π4A = \frac{\pi}{4}, B=π3B = \frac{\pi}{3}. cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin B.
10Problem 10
Answer
(a) θ=24−2rr\theta = \dfrac{24 - 2r}{r}. (b) A=r(12−r)=12r−r2A = r(12 - r) = 12r - r^2. (c) r=6r = 6, max area 36 cm2^2.
Full working
(a) Perimeter =2r+rθ= 2r + r\theta, so θ=24−2rr\theta = \frac{24 - 2r}{r}. (b) A=12r2θ=12r2⋅24−2rr=r(12−r)A = \frac{1}{2}r^2\theta = \frac{1}{2}r^2 \cdot \frac{24 - 2r}{r} = r(12 - r). (c) Quadratic A=−r2+12rA = -r^2 + 12r, max at r=6r = 6, A=36A = 36.
11Problem 11
Answer
(a) 33π≈10433\pi \approx 104 cm2^2. (b) 11π2≈17.3\frac{11\pi}{2} \approx 17.3 cm2^2. (c) 11π/3+6≈17.511\pi/3 + 6 \approx 17.5 cm. Wait — recompute: outer arc =7⋅π3=7π3= 7 \cdot \frac{\pi}{3} = \frac{7\pi}{3}; inner arc =4⋅π3=4π3= 4 \cdot \frac{\pi}{3} = \frac{4\pi}{3}; two radial edges total 2⋅3=62 \cdot 3 = 6. Total perimeter =7π+4π3+6=11π3+6≈17.5= \frac{7\pi + 4\pi}{3} + 6 = \frac{11\pi}{3} + 6 \approx 17.5 cm.
Full working
(a) Outer area =49π= 49\pi; inner area =16π= 16\pi; difference =33π= 33\pi. (b) Annular sector area =12(49−16)π3=33π6=11π2= \frac{1}{2}(49 - 16)\frac{\pi}{3} = \frac{33\pi}{6} = \frac{11\pi}{2}.
12Problem 12
Answer
(a) Both sides give 0. (b) Two more solutions visible, one positive, one negative. (c) x≈1.90x \approx 1.90. (d) The equation mixes a transcendental (sin⁡\sin) and a polynomial — there is no algebraic closed-form.
Full working
(a) sin⁡0=0\sin 0 = 0 and 0/2=00/2 = 0. (b) The line y=x/2y = x/2 has gradient 0.50.5; intersects sin⁡\sin once positive, once negative (by symmetry). (c) Use Newton's method or the GDC: x≈1.8955x \approx 1.8955. (d) Equations involving combinations of polynomial and transcendental functions are generally not solvable in closed form — only special cases (e.g. sin⁡x=0\sin x = 0) have explicit solutions.