Noesis
Level 2 — RecallAdvanced Trigonometry

Advanced Trigonometry

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Level: 2 (Recall / Standard Textbook Problems) Time Limit: 30 minutes Total Marks: 40

Q1. Convert the following. (4 marks) (a) 135°135° to radians. (b) 5π6\dfrac{5\pi}{6} radians to degrees.

Q2. A sector of a circle has radius 6 cm6\text{ cm} and subtends an angle of π3\dfrac{\pi}{3} radians at the centre. Find: (a) the arc length, (b) the area of the sector. (4 marks)

Q3. Using the ASTC rule and reference angles, evaluate exactly: (6 marks) (a) sin⁡210°\sin 210° (b) cos⁡135°\cos 135° (c) tan⁡300°\tan 300°

Q4. State the amplitude, period, phase shift and vertical shift of the function (4 marks) y=3sin⁡ ⁣(2x−π2)+1.y = 3\sin\!\left(2x - \frac{\pi}{2}\right) + 1.

Q5. Starting from sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, derive the identity 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta. (3 marks)

Q6. Using the sum formula, show that sin⁡(A+B)−sin⁡(A−B)=2cos⁡Asin⁡B\sin(A+B) - \sin(A-B) = 2\cos A\sin B. (4 marks)

Q7. Given cos⁡θ=35\cos\theta = \dfrac{3}{5} where θ\theta is in the first quadrant, find the exact values of sin⁡2θ\sin 2\theta and cos⁡2θ\cos 2\theta. (5 marks)

Q8. Solve for xx in the interval 0≤x<2π0 \le x < 2\pi: (4 marks) 2sin⁡x−1=0.2\sin x - 1 = 0.

Q9. In triangle ABCABC, a=8a = 8, b=5b = 5 and angle C=60°C = 60°. Find: (4 marks) (a) the length of side cc (Law of Cosines), (b) the area of the triangle.

Q10. State the domain and range of arcsin⁡x\arcsin x and arctan⁡x\arctan x. (2 marks)

End of Paper

Answer keyMark scheme & solutions

Q1. (4 marks) (a) Multiply by π180\frac{\pi}{180}: 135×π180=3π4135 \times \frac{\pi}{180} = \frac{3\pi}{4}. (2) (b) Multiply by 180π\frac{180}{\pi}: 5π6×180π=150°\frac{5\pi}{6}\times\frac{180}{\pi} = 150°. (2) Why: 180°=π180° = \pi rad is the conversion anchor.

Q2. (4 marks) (a) Arc length s=rθ=6×π3=2π≈6.28 cms = r\theta = 6\times\frac{\pi}{3} = 2\pi \approx 6.28\text{ cm}. (2) (b) Sector area =12r2θ=12(36)π3=6π≈18.85 cm2=\frac12 r^2\theta = \frac12(36)\frac{\pi}{3} = 6\pi \approx 18.85\text{ cm}^2. (2) Why: These formulas require θ\theta in radians.

Q3. (6 marks) (a) 210°210° in Q3, reference 30°30°; sine negative in Q3: sin⁡210°=−sin⁡30°=−12\sin210° = -\sin30° = -\frac12. (2) (b) 135°135° in Q2, reference 45°45°; cosine negative in Q2: cos⁡135°=−cos⁡45°=−22\cos135° = -\cos45° = -\frac{\sqrt2}{2}. (2) (c) 300°300° in Q4, reference 60°60°; tangent negative in Q4: tan⁡300°=−tan⁡60°=−3\tan300° = -\tan60° = -\sqrt3. (2)

Q4. (4 marks) Write as 3sin⁡ ⁣(2(x−π4))+13\sin\!\big(2(x-\frac{\pi}{4})\big)+1.

  • Amplitude =∣3∣=3=|3|=3 (1)
  • Period =2π2=π=\frac{2\pi}{2}=\pi (1)
  • Phase shift =π4=\frac{\pi}{4} to the right (1)
  • Vertical shift =+1=+1 (up) (1)

Q5. (3 marks) Start: sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1. (1) Divide every term by cos⁡2θ\cos^2\theta: sin⁡2θcos⁡2θ+1=1cos⁡2θ\frac{\sin^2\theta}{\cos^2\theta} + 1 = \frac{1}{\cos^2\theta}. (1) Hence tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta. (1)

Q6. (4 marks) sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B (1) sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B) = \sin A\cos B - \cos A\sin B (1) Subtract: sin⁡(A+B)−sin⁡(A−B)=2cos⁡Asin⁡B\sin(A+B)-\sin(A-B) = 2\cos A\sin B. (2)

Q7. (5 marks) cos⁡θ=35\cos\theta=\frac35, Q1 so sin⁡θ=45\sin\theta=\frac45. (1) sin⁡2θ=2sin⁡θcos⁡θ=2⋅45⋅35=2425\sin2\theta = 2\sin\theta\cos\theta = 2\cdot\frac45\cdot\frac35 = \frac{24}{25}. (2) cos⁡2θ=cos⁡2θ−sin⁡2θ=925−1625=−725\cos2\theta = \cos^2\theta-\sin^2\theta = \frac{9}{25}-\frac{16}{25} = -\frac{7}{25}. (2)

Q8. (4 marks) sin⁡x=12\sin x = \frac12. (1) Reference angle π6\frac{\pi}{6}; sine positive in Q1 and Q2. (1) x=π6x = \frac{\pi}{6} (1) and x=5π6x = \frac{5\pi}{6}. (1)

Q9. (4 marks) (a) c2=a2+b2−2abcos⁡C=64+25−2(8)(5)(0.5)=89−40=49c^2 = a^2+b^2-2ab\cos C = 64+25-2(8)(5)(0.5) = 89-40 = 49, so c=7c=7. (2) (b) Area =12absin⁡C=12(8)(5)sin⁡60°=20⋅32=103≈17.32=\frac12 ab\sin C = \frac12(8)(5)\sin60° = 20\cdot\frac{\sqrt3}{2} = 10\sqrt3 \approx 17.32. (2)

Q10. (2 marks) arcsin⁡x\arcsin x: domain [−1,1][-1,1], range [−π2,π2][-\frac{\pi}{2},\frac{\pi}{2}]. (1) arctan⁡x\arctan x: domain (−∞,∞)(-\infty,\infty), range (−π2,π2)(-\frac{\pi}{2},\frac{\pi}{2}). (1)

[
  {"claim":"135 degrees = 3pi/4 radians","code":"result = (Rational(135)*pi/180 == 3*pi/4)"},
  {"claim":"Sector arc length and area for r=6, theta=pi/3","code":"s = 6*(pi/3); A = Rational(1,2)*36*(pi/3); result = (s == 2*pi and A == 6*pi)"},
  {"claim":"sin2theta=24/25 and cos2theta=-7/25 for cos=3/5 Q1","code":"c=Rational(3,5); s=Rational(4,5); result = (2*s*c == Rational(24,25) and c**2 - s**2 == Rational(-7,25))"},
  {"claim":"Law of cosines gives c=7 and area=10sqrt3","code":"c2 = 64+25-2*8*5*Rational(1,2); area = Rational(1,2)*8*5*sin(pi/3); result = (c2 == 49 and sqrt(c2)==7 and simplify(area-10*sqrt(3))==0)"}
]