amazon.co.ukRadians, exact values, identities, compound and double angles, the R-formula, sine and cosine rules and trigonometric equations - A-Level pure maths trigonometry (Edexcel 9MA0 Topic 5, AQA 7357 Section E) revised with worked examples and mark-scheme logic.
A-level trigonometry is done in radians by default (Edexcel 9MA0 5.1-5.2; AQA 7357 E1-E2; OCR A H640 9.01; WJEC/Eduqas and CCEA AS Pure Mathematics): pi radians = 180 degrees, so pi/6 = 30 degrees, pi/4 = 45 degrees, pi/3 = 60 degrees, pi/2 = 90 degrees. Radians make arc length and sector area clean: arc length s = r(theta) and sector area A = 1/2 r^2 (theta), with theta in radians - no 360-degree factor needed. Convert always when a question mixes degrees and radians, and keep the calculator in the right mode: degree-mode calculators silently give wrong values in calculus questions.
Exact values are recall material, not derivation material. The sin row climbs 0, 1/2, sqrt(2)/2, sqrt(3)/2, 1 across 0, pi/6, pi/4, pi/3, pi/2; the cos row runs in reverse; tan = sin/cos gives 0, 1/sqrt(3), 1, sqrt(3), undefined. Answers demanding exact form must keep the surds - 0.866 loses the mark.
The graphs: y = sin x and y = cos x have amplitude 1 and period 2pi, with cos x = sin(x + pi/2); y = tan x has period pi and vertical asymptotes where cos x = 0, at pi/2 + k(pi). Transformations follow the general pattern y = a sin(bx + c) + d: amplitude |a|, period 2pi/|b|, phase shift -c/b, vertical shift d. So y = 2cos(2x - pi/3) has amplitude 2, period pi and phase shift pi/6 to the right - a max of 2 occurs at x = pi/6 + k(pi).
Inverse functions restrict to principal values: arcsin x has range [-pi/2, pi/2], arccos x has range [0, pi], arctan x has range (-pi/2, pi/2). These restrictions matter when solving equations involving inverse functions, since the inverse only returns one branch.
Two identities drive almost everything (Edexcel 9MA0 5.4; AQA 7357 E4; OCR A H640 10.01): sin^2(theta) + cos^2(theta) = 1 and tan(theta) = sin(theta)/cos(theta). The identity is used in two directions: to replace sin^2 with 1 - cos^2 (or vice versa) when an equation mixes both, and to prove identities. The solving routine: (1) isolate or factorise to get a single trig function equal to a constant; (2) find the reference (acute) angle; (3) use the CAST diagram or a graph to find all solutions in the interval; (4) check every solution lies in the required interval.
Solve 2cos^2(theta) + sin(theta) - 1 = 0 for 0 <= theta < 2pi, giving exact answers.
Line 1: Replace cos^2(theta) with 1 - sin^2(theta): 2(1 - sin^2(theta)) + sin(theta) - 1 = 0 (M1: identity substitution).
Line 2: -2sin^2(theta) + sin(theta) + 1 = 0, i.e. 2sin^2(theta) - sin(theta) - 1 = 0 (A1).
Line 3: Factorising: (2sin(theta) + 1)(sin(theta) - 1) = 0, so sin(theta) = -1/2 or sin(theta) = 1 (M1).
Line 4: sin(theta) = 1: theta = pi/2 (A1). sin(theta) = -1/2: reference angle pi/6, sine negative in quadrants 3 and 4, so theta = 7pi/6 and theta = 11pi/6 (A1 A1).
Two habits protect marks: never divide by a trig factor - cos(theta)(4sin(theta) - sqrt(3)) = 0 must be factorised, not divided by cos(theta), because dividing destroys the cos(theta) = 0 solutions; and always sketch or CAST before listing solutions, since the interval determines how many there are.
Solve 2sin(2theta) = sqrt(3)cos(theta) for 0 <= theta < 2pi.
Line 1: 2(2sin(theta)cos(theta)) = sqrt(3)cos(theta) (M1: double angle).
Line 2: 4sin(theta)cos(theta) - sqrt(3)cos(theta) = 0, so cos(theta)(4sin(theta) - sqrt(3)) = 0 (M1: factorise, do not divide).
Line 3: cos(theta) = 0: theta = pi/2, 3pi/2 (A1).
Line 4: sin(theta) = sqrt(3)/4 = 0.4330: theta = arcsin(0.4330) = 0.448 and theta = pi - 0.448 = 2.694 (A1 A1).
The compound angle formulae (Edexcel 9MA0 5.5; AQA 7357 E6; OCR A H640 10.01) expand sin(A +/- B), cos(A +/- B) and tan(A +/- B); the double angle formulae follow by setting B = A. sin 2A = 2 sin A cos A; cos 2A = cos^2 A - sin^2 A = 2cos^2 A - 1 = 1 - 2sin^2 A (three interchangeable forms - choose the one matching the rest of the equation); tan 2A = 2 tan A/(1 - tan^2 A), derived by substituting A = B into the tan compound formula and using tan(theta) = sin(theta)/cos(theta).
The R-formula converts a sin(theta) + b cos(theta) into a single shifted wave: a sin(theta) + b cos(theta) = R sin(theta + alpha) with R = sqrt(a^2 + b^2) and tan(alpha) = b/a. The payoffs: the maximum of the expression is R, the minimum is -R, and equations of the form a sin(theta) + b cos(theta) = c become one-step inverse sine problems. For a cos form, use R cos(theta - alpha) with tan(alpha) = b/a.
Write 5cos(theta) - 12sin(theta) in the form R cos(theta + alpha) with R > 0 and 0 < alpha < pi/2, and hence find the maximum value.
Line 1: R cos(theta + alpha) = R cos(theta)cos(alpha) - R sin(theta)sin(alpha) (M1: expand).
Line 2: Compare: R cos(alpha) = 5 and R sin(alpha) = 12 (M1).
Line 3: R^2 = 5^2 + 12^2 = 169, so R = 13 (A1). tan(alpha) = 12/5, so alpha = arctan(12/5) = 1.176 radians (A1).
Line 4: 5cos(theta) - 12sin(theta) = 13cos(theta + 1.176); maximum of cos is 1, so the maximum value is 13 (A1).
Know the direction of the shift: R sin(theta + alpha) shifts the sine wave left by alpha; using the wrong sign flips the quadrant of alpha and loses the accuracy marks. When a question asks where the maximum occurs, set theta + alpha = pi/2 and solve.
Non-right-angled triangles (Edexcel 9MA0 5.3; AQA 7357 E3; OCR A H640 9.01; CCEA AS1/A21) use three results. The sine rule a/sin A = b/sin B = c/sin C applies when you know two angles and a side, or two sides and a non-included angle - the latter is the ambiguous case, where two triangles can fit the data. The cosine rule a^2 = b^2 + c^2 - 2bc cos A applies when two sides and the included angle are known, or all three sides (to find an angle by rearranging to cos A = (b^2 + c^2 - a^2)/2bc). The area is 1/2 ab sin C for any pair of sides and the included angle.
In triangle ABC, AB = 8 cm, BC = 11 cm and angle ABC = 72 degrees. Find AC and the area.
Line 1: The included angle at B sits between AB and BC: AC^2 = 8^2 + 11^2 - 2(8)(11)cos(72 degrees) (M1: cosine rule).
Line 2: = 64 + 121 - 176(0.3090) = 185 - 54.4 = 130.6 (M1).
Line 3: AC = sqrt(130.6) = 11.4 cm (A1: 3 s.f.).
Line 4: Area = 1/2 (8)(11)sin(72 degrees) = 44(0.9511) = 41.8 cm^2 (A1).
In triangle ABC, a = 7, b = 9 and angle C = 60 degrees. Find c and the area.
Line 1: The angle is included between a and b, so use the cosine rule: c^2 = 7^2 + 9^2 - 2(7)(9)cos(60 degrees) = 49 + 81 - 63 = 67 (M1 A1).
Line 2: c = sqrt(67) = 8.19 (A1).
Line 3: Area = 1/2 ab sin C = 1/2 (7)(9)sin(60 degrees) = 31.5 x (sqrt(3)/2) = 27.3 (A1).
Ambiguous-case discipline: with the sine rule in the form sin B = b sin A/a, the acute solution for B may not be the only one - the obtuse supplement 180 degrees - B is also valid provided the angles still sum below 180 degrees. State both when the data allows, or justify the single solution.
Identity proofs follow one discipline: start from the more complicated side and reduce it to the other, never moving terms across the equals sign as if solving. The universal strategy: convert everything to sin and cos using tan(theta) = sin(theta)/cos(theta), then look for the sin^2 + cos^2 = 1 substitution.
Prove that tan(2theta) = 2tan(theta)/(1 - tan^2(theta)).
Line 1: sin(2theta) = 2sin(theta)cos(theta) and cos(2theta) = cos^2(theta) - sin^2(theta) (M1: double angles).
Line 2: tan(2theta) = sin(2theta)/cos(2theta) = 2sin(theta)cos(theta)/(cos^2(theta) - sin^2(theta)) (M1: tan as sin/cos).
Line 3: Divide numerator and denominator by cos^2(theta): 2tan(theta)/(1 - tan^2(theta)) (A1: reduction complete, valid where cos(theta) is not 0).
Exact values with inverse functions combine the ideas: sin(arcsin(3/5) + arccos(4/5)) - both inverse values equal the same angle (36.87 degrees, since sin = 3/5 and cos = 4/5 there), so the expression is sin(2 x 36.87 degrees) = 2(3/5)(4/5) = 24/25. Convert inverse values to a right triangle, then apply a compound or double angle formula.
Exam technique: (1) keep the calculator in the mode the question uses - radians unless degrees are stated; (2) never divide by a trig factor; (3) give exact surd answers unless decimals are requested, and give decimals to 3 s.f. otherwise; (4) check all solutions lie in the interval, including the boundary behaviour at pi/2 asymptotes; (5) for R-formula questions, state R and alpha explicitly before using them. Small-angle approximations (sin(theta) = theta, cos(theta) = 1 - theta^2/2, tan(theta) = theta for small theta in radians) appear in derivation questions - know them. Board coverage: Edexcel 9MA0 Topic 5, AQA 7357 E1-E6, OCR A H640 9.01 and 10.01, WJEC/Eduqas AS Unit 1 trigonometry, CCEA AS1/A21 - identical techniques, differing wording.