Grade 12 — Term 2
7 study notes · 126 practice questions · Aligned with NIE syllabus
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This term in the Idasara Academy app has 7 lessons (NIE Units 9, 11, 13, 14, 15). For each lesson, this page gives you the points you must know and one exam-style question with its answer. None of it needs an account.
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Lessons in This Term
Circular (trigonometric) functions
What you must know
- Product-to-Sum and Sum-to-Product conversions: 2 sin A cos B = sin(A + B) + sin(A - B); 2 cos A cos B = cos(A + B) + cos(A - B); 2 sin A sin B = cos(A - B) - cos(A + B); sin C + sin D = 2 sin((C+D)/2) cos((C-D)/2); cos C - cos D = -2 sin((C+D)/2) sin((C-D)/2).
- t-substitution (Half-angle tangent substitution): Let t = tan(θ/2); then sin θ = 2t/(1 + t^2), cos θ = (1 - t^2)/(1 + t^2), and tan θ = 2t/(1 - t^2); converts trigonometric equations into algebra.
- General solutions of trigonometric equations: sin θ = sin α => θ = nπ + (-1)^n α; cos θ = cos α => θ = 2nπ ± α; tan θ = tan α => θ = nπ + α (where n ∈ ℤ).
Try one question
Which of the following trigonometric identities expresses sin(2θ) in terms of tan(θ)?
- A. (2 tan θ) / (1 - tan² θ)
- B. (2 tan θ) / (1 + tan² θ)
- C. (1 - tan² θ) / (1 + tan² θ)
- D. (tan θ) / (1 + tan² θ)
Show the answer
Answer: B. (2 tan θ) / (1 + tan² θ). Starting from the double angle identity sin(2θ) = 2 sin θ cos θ = 2 (sin θ / cos θ) cos² θ = 2 tan θ / sec² θ. Since sec² θ = 1 + tan² θ, we obtain sin(2θ) = (2 tan θ) / (1 + tan² θ).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Sine rule and cosine rule
What you must know
- Ambiguous Case of Sine Rule (SSA): Given sides a, b and non-included angle A (with A acute): if a < b sin A, no triangle exists; if a = b sin A, exactly one right-angled triangle exists; if b sin A < a < b, TWO triangles exist; if a ≥ b, exactly one triangle exists.
- Inradius and Circumradius formulas: r = Δ/s = (s - a)tan(A/2); R = abc/(4Δ) = a/(2 sin A); connection: r = 4R sin(A/2) sin(B/2) sin(C/2).
- Half-angle tangent formula (Napier's Analogies): tan((B - C)/2) = [(b - c)/(b + c)] cot(A/2).
Try one question
In a triangle ABC, sides b = 3 cm, c = 5 cm, and angle A = 60°. Using the Cosine Rule, what is the exact length of side a?
- A. √34 cm
- B. 4 cm
- C. √19 cm
- D. √49 cm (7 cm)
Show the answer
Answer: C. √19 cm. By the Cosine Rule: a² = b² + c² - 2bc cos A. Substituting b = 3, c = 5, A = 60° (cos 60° = 1/2): a² = 3² + 5² - 2(3)(5)(1/2) = 9 + 25 - 15 = 19. Hence, a = √19 cm.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Sine rule and cosine rule (Pure Maths)
What you must know
- Apollonius' Theorem proof via Cosine Rule: Apply cosine rule to triangles ABD and ACD where cos(∠ADB) = -cos(∠ADC) because ∠ADB + ∠ADC = 180°; adding the two equations cancels the cross term to give AB^2 + AC^2 = 2(AD^2 + BD^2).
- Exradius formulas: ra = Δ / (s - a) = s tan(A/2) = 4R sin(A/2) cos(B/2) cos(C/2); key identity: 1/ra + 1/rb + 1/rc = 1/r.
- Three-dimensional angle determination: When finding angle between a line and a plane, always drop a perpendicular from a point on the line to the plane, identifying the orthogonal projection.
Try one question
In any triangle ABC with circumradius R and standard notation, which of the following expresses the area Δ of the triangle?
- A. Δ = (a b c) / (4R)
- B. Δ = 2 R² sin A sin B sin C
- C. Δ = (a + b + c) / (2R)
- D. Δ = (a b c) / (2R)
Show the answer
Answer: A. Δ = (a b c) / (4R). Area of triangle Δ = ½ a b sin C. From the Sine Rule, c / sin C = 2R => sin C = c / (2R). Substituting gives Δ = ½ a b [c / (2R)] = (a b c) / (4R).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Limits of a function
What you must know
- Sandwich / Squeeze Theorem: If g(x) ≤ f(x) ≤ h(x) in an open interval around a, and lim_{x->a} g(x) = lim_{x->a} h(x) = L, then lim_{x->a} f(x) = L; foundational proof for lim_{x->0} (sin x)/x = 1 using unit circle geometric areas.
- Limits at infinity for rational functions: For P(x)/Q(x) as x -> ∞: if deg(P) < deg(Q), limit is 0; if deg(P) = deg(Q), limit is ratio of leading coefficients; if deg(P) > deg(Q), limit diverges to ±∞.
- Radian requirement pitfall: Formula lim_{x->0} (sin x)/x = 1 holds ONLY when x is measured in radians; if x is in degrees, the limit is π/180.
Try one question
What is the value of the limit lim_{x → 0} [sin(3x) / x]?
- A. 0
- B. 3
- C. 1
- D. 1/3
Show the answer
Answer: B. 3. Rewriting the expression: lim_{x → 0} [sin(3x) / x] = 3 × lim_{x → 0} [sin(3x) / (3x)]. Since lim_{u → 0} [sin(u) / u] = 1, this equals 3 × 1 = 3.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Differentiation techniques
What you must know
- Second derivative of parametric function pitfall: d^2y/dx^2 is NOT (d^2y/dt^2)/(d^2x/dt^2); it MUST be computed as d/dt(dy/dx) divided by dx/dt.
- Logarithmic differentiation technique: For functions of the form y = [f(x)]^g(x) or complex products/quotients, take ln on both sides: ln y = g(x) * ln f(x), then differentiate implicitly.
- Tangents and Normals: Gradient of tangent is m = (dy/dx)|_{(x1, y1)}; equation of tangent is y - y1 = m(x - x1); gradient of normal is -1/m; equation of normal is y - y1 = (-1/m)(x - x1).
Try one question
What is the derivative with respect to x of the function y = x² · sin(x)?
- A. 2x cos(x)
- B. x² cos(x) - 2x sin(x)
- C. 2x sin(x) + x² cos(x)
- D. 2x sin(x) - x² cos(x)
Show the answer
Answer: C. 2x sin(x) + x² cos(x). Applying the Product Rule for differentiation d/dx (u · v) = u' v + u v' with u = x² and v = sin(x): dy/dx = (2x) · sin(x) + x² · cos(x) = 2x sin(x) + x² cos(x).
A model question written by Idasara on the exam pattern. It is not from a past paper.
The definite integral and the Fundamental Theorem
What you must know
- Leibniz Integral Rule for differentiating an integral: d/dx [∫_{u(x)}^{v(x)} f(t) dt] = f(v(x)) * v'(x) - f(u(x)) * u'(x).
- Integration by Parts for definite integrals: ∫_a^b u v' dx = [u v]_a^b - ∫_a^b v u' dx; select u using LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential).
- Periodic function property: If f(x) has period T (f(x + T) = f(x)), then ∫_0^{nT} f(x) dx = n ∫_0^T f(x) dx for any integer n.
Try one question
What is the exact value of the definite integral ∫₀^(π/2) sin(x) dx?
- A. π / 2
- B. 0
- C. 1
- D. -1
Show the answer
Answer: C. 1. The anti-derivative of sin(x) is -cos(x). By the Fundamental Theorem of Calculus: ∫₀^(π/2) sin(x) dx = [-cos(x)]₀^(π/2) = -cos(π/2) - (-cos(0)) = 0 - (-1) = 1.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Area bounded by curves
What you must know
- Sign trap in definite area integrals: Never compute ∫_a^b f(x) dx directly across an x-intercept where f(x) changes sign; positive and negative regions cancel algebraically, understating physical geometric area.
- Points of intersection as integration limits: Set f(x) = g(x) to solve for intersection coordinates; always sketch the curves to identify which curve lies above the other.
- Symmetry simplification: If the region is symmetric across an axis (e.g., ellipse x^2/a^2 + y^2/b^2 = 1), compute area of one quadrant and multiply by 4: Total Area = 4 * ∫_0^a (b/a)√(a^2 - x^2) dx = πab.
Try one question
What is the area of the finite plane region enclosed between the parabola y = x² and the straight line y = x?
- A. 1/6
- B. 5/6
- C. 1/3
- D. 1/2
Show the answer
Answer: A. 1/6. Intersection points: x² = x => x(x - 1) = 0 => x = 0 and x = 1. On [0, 1], the line lies above the parabola (x ≥ x²). Area = ∫₀¹ (x - x²) dx = [x²/2 - x³/3]₀¹ = (1/2 - 1/3) - 0 = 1/6.
A model question written by Idasara on the exam pattern. It is not from a past paper.
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