Topic 5 is the biggest pure topic. This mini-lesson covers radians (arc length and sector area), exact values, the reciprocal functions sec, cosec and cot, the inverse functions and their restricted domains, the identities (Pythagorean, compound-angle, double-angle), the small-angle approximations, and the harmonic form R cos(x ± α) / R sin(x ± α).
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Watch for the Exam trap notes — that is where the marks get thrown away. Press Start when you are ready.
One radian is the angle subtended at the centre by an arc equal in length to the radius.
Arc length s = rθ = 8 × 0.75 = 6 cm.
Sector area A = ½r²θ = ½ × 64 × 0.75 = 32 × 0.75 = 24 cm².
Segment area = sector − triangle = ½r²θ − ½r² sin θ = 24 − 32 sin(0.75) = 24 − 21.81 = 2.19 cm².
Exam trap: if your calculator is in degrees, sin(0.75) gives 0.0131 instead of 0.6816 and the whole question collapses. Set it to RAD and leave it there.
s = rθ = 8 × 0.75 = 6 cm.
A = ½ × 64 × 0.75 = 32 × 0.75 = 24 cm².
You must know these without a calculator:
The reciprocal trigonometric functions:
The inverse functions need restricted domains to be one-to-one: arcsin has range [−π/2, π/2], arccos has range [0, π], arctan has range (−π/2, π/2).
Exam trap: sec θ is not arccos θ. sec θ = 1/cos θ; arccos θ is the angle whose cosine is θ. Different beasts entirely.
Everything starts with one identity and its two derived forms.
Replace sec²θ with 1 + tan²θ: 2(1 + tan²θ) = 5 + 3 tan θ
2tan²θ − 3tan θ − 3 = 0 — a quadratic in tan θ, which you can now solve.
Exam trap: when you divide an equation by cos θ or sin θ you lose roots. Factorise instead: sin θ cos θ = sin θ ⇒ sin θ(cos θ − 1) = 0 gives sin θ = 0 and cos θ = 1. Dividing by sin θ would have thrown away the first family.
Put B = A to get the double-angle formulae:
cos 2θ = 1 − 2sin²θ = 1 − 2(9/25) = 1 − 18/25 = 7/25 = 0.28.
Check the long way: cos θ = 4/5 (3-4-5 triangle), so cos²θ − sin²θ = 16/25 − 9/25 = 7/25 ✓
Why three versions of cos 2A? Pick the one that matches what you have. For integration you want the rearrangements cos²A ≡ ½(1 + cos 2A) and sin²A ≡ ½(1 − cos 2A) — these turn a square into something integrable.
sin²θ = 9/25.
cos 2θ = 1 − 2(9/25) = 1 − 18/25 = 7/25 = 0.28.
Tap a statement, then decide: always true, only true for particular θ, or simply a wrong identity.
When θ is small and measured in radians:
Numerator: 1 − cos 2θ ≈ 1 − [1 − (2θ)²/2] = 1 − 1 + 2θ² = 2θ².
Denominator: θ sin θ ≈ θ × θ = θ².
Ratio ≈ 2θ²/θ² = 2.
Numerical check at θ = 0.01: (1 − cos 0.02)/(0.01 × sin 0.01) = 0.00019999/0.00009999… = 2.000 ✓
Exam trap: apply the approximation to the whole angle. In cos 2θ the angle is 2θ, so it becomes 1 − (2θ)²/2 = 1 − 2θ², not 1 − θ²/2.
cos 2θ ≈ 1 − (2θ)²/2 = 1 − 2θ², so 1 − cos 2θ ≈ 2θ².
sin θ ≈ θ, so θ sin θ ≈ θ².
Ratio ≈ 2θ² ÷ θ² = 2.
Any a sin x + b cos x can be rewritten as a single sine or cosine wave. That instantly gives the maximum, the minimum and every solution.
Expand: R sin(x + α) = R sin x cos α + R cos x sin α.
Compare coefficients: R cos α = 3 and R sin α = 4.
Square and add: R² (cos²α + sin²α) = 9 + 16 = 25 ⇒ R = 5.
Divide: tan α = 4/3 ⇒ α = 53.13° (0.927 rad).
So 3 sin x + 4 cos x ≡ 5 sin(x + 53.13°). Check at x = 0: LHS = 4; RHS = 5 sin 53.13° = 5 × 0.8 = 4 ✓
Maximum value 5 (when sin = 1); minimum −5.
Exam trap: match the form asked for. For R cos(x − α) you compare with R cos x cos α + R sin x sin α, so the roles of a and b swap and tan α = a/b. Always expand the requested form first, then compare.
R = √(9 + 16) = √25 = 5. This is also the maximum value of the expression.
tan α = 4/3 ⇒ α = arctan(1.3333) = 53.130…° = 53.1° (1 d.p.).
Check: 5 sin(0 + 53.13°) = 5 × 0.8 = 4 = the value of 3 sin 0 + 4 cos 0 ✓
Tap a card on the left, then its definition or value.
Proof link: compound-angle formulae are also proved and applied in Topic 1-style questions — e.g. show that (1 + cos 2θ)/sin 2θ ≡ cot θ. Work on one side only and quote the identities you use.
Radians: π = 180° · s = rθ · A = ½r²θ · calculator in RAD
Identities: sin²+cos² ≡ 1 · 1 + tan² ≡ sec² · 1 + cot² ≡ cosec²
Double angle: sin 2A ≡ 2 sin A cos A · cos 2A ≡ 2cos²A − 1 ≡ 1 − 2sin²A
Small angles: sin θ ≈ θ · tan θ ≈ θ · cos θ ≈ 1 − θ²/2 (radians only)
Harmonic form: a sin x + b cos x ≡ R sin(x + α), R = √(a² + b²), max R, min −R
Solving: widen the interval, never divide by a trig function, factorise
Topic 5 is the topic that feeds differentiation, integration and modelling — the identities have to be automatic. Press Finish to see your score.
You have worked through Trigonometry for Edexcel A-level Mathematics (9MA0). 🎉
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