This mini-lesson covers AQA section E — Trigonometry: radians (arc length and sector area), small-angle approximations, the reciprocal functions sec, cosec and cot, the inverse functions and their restricted ranges, the Pythagorean identities, compound and double angle formulae, and the harmonic form R sin(x ± α) for solving equations and finding maxima.
Work through each screen, answer the questions as you go (a few are conceptual, most are calculations) and collect ⭐ stars. Everything here is A-level standard, drawn from the AQA 7357 subject content. Press Start when you are ready.
A radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. This makes the formulae beautifully simple — but only if θ is in radians.
Common conversions: π/6 = 30° · π/4 = 45° · π/3 = 60° · π/2 = 90° · 2π = 360°.
A sector has radius r = 8 cm and angle θ = 0.75 rad.
Arc length s = rθ = 8 × 0.75 = 6 cm.
Sector area A = ½r²θ = ½ × 64 × 0.75 = 24 cm².
Perimeter of the sector = arc + two radii = 6 + 8 + 8 = 22 cm.
Check your calculator mode. Working in degrees when the question is in radians is the single biggest source of lost marks in this topic.
When θ is small and measured in radians, the trig functions are well approximated by simple polynomials:
Check cos: cos(0.02) ≈ 1 − (0.02)²/2 = 1 − 0.0002 = 0.9998. The true value is 0.99980001 — accurate to 7 decimal places.
Estimate (2 sin θ + tan θ)/θ for small θ.
Replace sin θ ≈ θ and tan θ ≈ θ: (2θ + θ)/θ = 3θ/θ = 3.
The reason it matters: sin θ ≈ θ is exactly what makes d/dx(sin x) = cos x work in the first-principles proof. These approximations are not a trick — they underpin the calculus of trig functions.
Three reciprocal functions, and a warning about notation.
The inverse functions arcsin, arccos and arctan (also written sin⁻¹ etc.) undo the trig functions — but each needs a restricted domain to be one-to-one, which forces a restricted range:
Notation trap: sin⁻¹x means the inverse function, whereas sin²x means (sin x)². So sin⁻¹x is not 1/sin x — that is cosec x.
Tap an identity, then tap the family it belongs to.
One identity, and two consequences that you must be able to derive on the spot.
Also essential: tan θ ≡ sin θ / cos θ. Between them, these convert almost any expression into one function.
Solve 2 sin²θ + 3 cos θ = 3 for 0 ≤ θ < 2π.
Use sin²θ = 1 − cos²θ: 2(1 − cos²θ) + 3cos θ = 3 ⇒ −2cos²θ + 3cos θ − 1 = 0 ⇒ 2cos²θ − 3cos θ + 1 = 0.
Factorise: (2cos θ − 1)(cos θ − 1) = 0 ⇒ cos θ = ½ or cos θ = 1.
So θ = π/3, 5π/3, 0.
The strategy is always the same: use an identity to get everything in terms of one trig function, then treat it as an ordinary quadratic.
These are given in the formula book, but you must know how to choose and use them.
Note the sign flip in the cosine formula: cos(A + B) has a minus in the middle. This catches people out every year.
cos 75° = cos(45° + 30°) = cos45 cos30 − sin45 sin30
= (√2/2)(√3/2) − (√2/2)(1/2) = (√6 − √2)/4.
Not distributive! sin(A + B) is not sin A + sin B. Test it: sin(90°) = 1, but sin 45° + sin 45° = √2 ≈ 1.41.
Put B = A in the compound angle formulae and the double angle formulae drop out:
The three versions of cos 2θ are the point of the topic — pick the one that leaves you with the function you want:
Careful: sin 2θ ≠ 2 sin θ. Try θ = 30°: sin 60° = 0.866, but 2 sin 30° = 1.
Any expression a sin x + b cos x can be squeezed into a single trig function — which immediately gives you its maximum, its minimum, and an easy route to solving the equation.
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 = 3² + 4² = 25, so R = 5.
Divide: tan α = 4/3, so α = 53.1° (1 d.p.), or 0.927 rad.
So 3 sin x + 4 cos x ≡ 5 sin(x + 53.1°).
Why this is worth doing: the maximum of 5 sin(x + α) is 5 and its minimum is −5, read straight off R. And 3 sin x + 4 cos x = 2 becomes sin(x + 53.1°) = 0.4 — a routine equation.
Tap an item on the left, then its partner on the right.
The last mark is nearly always for finding all the solutions in the given interval. A reliable procedure:
Never divide by a trig function. From 2 sin x cos x = sin x, cancelling sin x destroys the solutions where sin x = 0. Factorise instead: sin x(2cos x − 1) = 0.
Radians: π rad = 180° · s = rθ · A = ½r²θ (θ in radians)
Small angles: sin θ ≈ θ · tan θ ≈ θ · cos θ ≈ 1 − θ²/2
Reciprocals: sec = 1/cos · cosec = 1/sin · cot = cos/sin
Identities: sin² + cos² ≡ 1 ⇒ 1 + tan² ≡ sec² and 1 + cot² ≡ cosec²
Double angle: sin 2θ = 2 sin θ cos θ · cos 2θ has three forms — choose the useful one
Harmonic form: a sin x + b cos x = R sin(x + α), R = √(a² + b²) — max is R, min is −R
Equations: stretch the interval first, then use symmetry to find every solution
That is AQA section E — Trigonometry. Press Finish to see your score.
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