Noesis
Interleaved — Phase 3

Maths interleaved practice

printable — key stays hidden on paper

Instructions: Work each problem on its own. These problems are deliberately mixed across many subtopics — decide which tool applies before computing. Show all working. Give exact values unless a decimal is requested. Total: 50 marks.

1. A sector of a circle has radius 66 cm and subtends an angle of 75°75° at the centre. Find (a) the arc length and (b) the sector area. Give answers to 3 significant figures. [5]

2. Without a calculator, evaluate cos⁡210°\cos 210° and tan⁡300°\tan 300°, stating the reference angle and the sign (via ASTC) in each case. [5]

3. Solve 2sin⁡2x−sin⁡x−1=02\sin^2 x - \sin x - 1 = 0 for x∈[0°,360°)x \in [0°, 360°). [6]

4. Simplify cos⁡(90°−θ)sin⁡(90°−θ)\dfrac{\cos(90° - \theta)}{\sin(90° - \theta)} and state the single trig function it equals. [3]

5. The function y=3sin⁡ ⁣(2x−π3)+1y = 3\sin\!\left(2x - \dfrac{\pi}{3}\right) + 1 is given. State its amplitude, period, phase shift, and range. [5]

6. Using the exact value cos⁡15°\cos 15°, evaluate it via a difference formula and give the exact surd form. [4]

7. In triangle ABCABC, a=8a = 8, b=5b = 5, and angle C=60°C = 60°. Find side cc and the area of the triangle. [6]

8. Given sin⁡θ=35\sin\theta = \dfrac{3}{5} with θ\theta in the second quadrant, find cos⁡2θ\cos 2\theta using an appropriate double-angle form. [4]

9. Simplify (a3b−2a−1b4)2\left(\dfrac{a^{3}b^{-2}}{a^{-1}b^{4}}\right)^{2} using the laws of exponents, leaving no negative indices. [4]

10. Express sin⁡75°cos⁡15°\sin 75° \cos 15° as a sum using a product-to-sum formula, then evaluate exactly. [4]

11. Evaluate arcsin⁡ ⁣(−12)+arccos⁡ ⁣(−12)\arcsin\!\left(-\dfrac{1}{2}\right) + \arccos\!\left(-\dfrac{1}{2}\right), giving the answer in radians, and justify using the ranges of the inverse functions. [4]

Answer keyMark scheme & solutions

1. Tests 3.1.3 (arc length & sector area) with 3.1.2 (conversion). Why: "radius + angle" → convert to radians first, then use s=rθs=r\theta, A=12r2θA=\tfrac12 r^2\theta. θ=75°=75⋅π180=5π12≈1.3090\theta = 75° = 75\cdot\frac{\pi}{180} = \frac{5\pi}{12} \approx 1.3090 rad. (a) s=rθ=6(1.3090)=7.854≈7.85s = r\theta = 6(1.3090) = 7.854 \approx \mathbf{7.85} cm. (b) A=12r2θ=12(36)(1.3090)=23.56≈23.6A = \tfrac12 r^2\theta = \tfrac12(36)(1.3090) = 23.56 \approx \mathbf{23.6} cm².

2. Tests 3.1.4 (ASTC) + 3.1.5 (reference angles). Why: angles > 90° → find reference angle + apply ASTC sign. cos⁡210°\cos 210°: Q3, reference =210−180=30°= 210-180 = 30°. In Q3 cosine is negative. cos⁡210°=−cos⁡30°=−32\cos 210° = -\cos 30° = \mathbf{-\frac{\sqrt3}{2}}. tan⁡300°\tan 300°: Q4, reference =360−300=60°= 360-300 = 60°. In Q4 tangent is negative. tan⁡300°=−tan⁡60°=−3\tan 300° = -\tan 60° = \mathbf{-\sqrt3}.

3. Tests 3.1.18 (solving trig equations). Why: quadratic in sin⁡x\sin x → factor, then solve each linear equation over the range. Let s=sin⁡xs=\sin x: 2s2−s−1=(2s+1)(s−1)=0⇒s=−122s^2 - s - 1 = (2s+1)(s-1)=0 \Rightarrow s = -\tfrac12 or s=1s=1. sin⁡x=1⇒x=90°\sin x = 1 \Rightarrow x = 90°. sin⁡x=−12⇒\sin x = -\tfrac12 \Rightarrow reference 30°30°, Q3 & Q4: x=210°,330°x = 210°, 330°. Solutions: x=90°, 210°, 330°\mathbf{x = 90°,\ 210°,\ 330°}.

4. Tests 3.1.11 (co-function) + 3.1.10 (quotient identity). Why: 90°−θ90°-\theta signals co-function conversion, then a quotient. cos⁡(90°−θ)=sin⁡θ\cos(90°-\theta)=\sin\theta, sin⁡(90°−θ)=cos⁡θ\sin(90°-\theta)=\cos\theta. sin⁡θcos⁡θ=tan⁡θ\dfrac{\sin\theta}{\cos\theta} = \mathbf{\tan\theta}.

5. Tests 3.1.8 (transformations of trig graphs). Why: read off A,B,C,DA,B,C,D from Asin⁡(Bx+C)+DA\sin(Bx+C)+D. Amplitude =∣A∣=3=|A|=\mathbf{3}. Period =2π∣B∣=2π2=π=\frac{2\pi}{|B|}=\frac{2\pi}{2}=\mathbf{\pi}. Phase shift: solve 2x−π3=0⇒x=π62x-\frac{\pi}{3}=0 \Rightarrow x=\frac{\pi}{6}, shift π6\mathbf{\frac{\pi}{6}} right. Range: centre D=1D=1, spread ±3\pm3 → [−2, 4]\mathbf{[-2,\,4]}.

6. Tests 3.1.12 (difference formula). Why: 15°=45°−30°15° = 45°-30° → use cos⁡(A−B)\cos(A-B). cos⁡15°=cos⁡(45°−30°)=cos⁡45cos⁡30+sin⁡45sin⁡30\cos 15° = \cos(45°-30°) = \cos45\cos30 + \sin45\sin30 =22⋅32+22⋅12=6+24=6+24= \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} + \frac{\sqrt2}{2}\cdot\frac12 = \frac{\sqrt6+\sqrt2}{4} = \mathbf{\frac{\sqrt6+\sqrt2}{4}}.

7. Tests 3.1.20 (law of cosines) + 3.1.21 (area = ½ab sin C). Why: two sides + included angle → cosine rule for cc, then 12absin⁡C\tfrac12 ab\sin C for area. c2=a2+b2−2abcos⁡C=64+25−2(8)(5)(12)=89−40=49⇒c=7c^2 = a^2+b^2-2ab\cos C = 64+25-2(8)(5)(\tfrac12) = 89-40 = 49 \Rightarrow c=\mathbf{7}. Area =12(8)(5)sin⁡60°=20⋅32=103≈17.3=\tfrac12(8)(5)\sin60° = 20\cdot\frac{\sqrt3}{2} = \mathbf{10\sqrt3} \approx 17.3.

8. Tests 3.1.13 (double angle) + 3.1.1 (quadrant sign). Why: only sin⁡θ\sin\theta needed → use cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\theta (avoids needing cos⁡θ\cos\theta's sign). cos⁡2θ=1−2(35)2=1−2⋅925=1−1825=725\cos 2\theta = 1 - 2(\tfrac35)^2 = 1 - 2\cdot\tfrac{9}{25} = 1 - \tfrac{18}{25} = \mathbf{\frac{7}{25}}.

9. Tests 3.2.2 (laws of exponents). Why: quotient → subtract exponents, then raise to power. Inside: a3−(−1)b−2−4=a4b−6a^{3-(-1)}b^{-2-4} = a^4 b^{-6}. Square: a8b−12=a8b12a^{8}b^{-12} = \mathbf{\dfrac{a^{8}}{b^{12}}}.

10. Tests 3.1.15 (product-to-sum). Why: product sin⁡cos⁡\sin\cos → sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)]\sin A\cos B = \tfrac12[\sin(A+B)+\sin(A-B)]. sin⁡75cos⁡15=12[sin⁡90°+sin⁡60°]=12[1+32]=2+34\sin75\cos15 = \tfrac12[\sin90° + \sin60°] = \tfrac12\left[1 + \tfrac{\sqrt3}{2}\right] = \mathbf{\frac{2+\sqrt3}{4}}.

11. Tests 3.1.17 (inverse trig domain/range). Why: each inverse has a fixed principal range — read the correct branch. arcsin⁡(−12)=−π6\arcsin(-\tfrac12) = -\frac{\pi}{6} (range [−π2,π2][-\frac\pi2,\frac\pi2]). arccos⁡(−12)=2π3\arccos(-\tfrac12) = \frac{2\pi}{3} (range [0,π][0,\pi]). Sum =−π6+2π3=−π6+4π6=π2= -\frac{\pi}{6} + \frac{2\pi}{3} = -\frac{\pi}{6}+\frac{4\pi}{6} = \mathbf{\frac{\pi}{2}}.

[
  {"claim":"cos(2θ)=7/25 when sinθ=3/5","code":"import sympy as sp\ns=sp.Rational(3,5)\nresult=(1-2*s**2)==sp.Rational(7,25)"},
  {"claim":"law of cosines gives c=7","code":"import sympy as sp\nc2=8**2+5**2-2*8*5*sp.Rational(1,2)\nresult=sp.sqrt(c2)==7"},
  {"claim":"arcsin(-1/2)+arccos(-1/2)=pi/2","code":"import sympy as sp\nval=sp.asin(sp.Rational(-1,2))+sp.acos(sp.Rational(-1,2))\nresult=sp.simplify(val-sp.pi/2)==0"}
]