Trigonometry (Pure 3)

• Sec goes with cos, cosec with sin. \(\cos(A + B)\) has a minus sign. Factorise, never divide. \(R = \sqrt{a^2 + b^2}\), and shift the interval before you solve.
• Sec with cos, cosec with sin. \(\cos(A + B)\) has a minus sign. Factorise, never divide. \(R\cos\alpha = a\), \(R\sin\alpha = b\), then shift the interval.

• 3.3.1 2.3.1 I can understand — understand the relationship of the secant, cosecant and cotangent functions to cosine, sine and tangent, and use properties and graphs of all six trigonometric functions for angles of any magnitude
• 3.3.2 2.3.2 I can use — use trigonometrical identities for the simplification and exact evaluation of expressions, and in the course of solving equations, and select an identity or identities appropriate to the context, showing familiarity in particular with the use of (a) sec²θ ≡ 1 + tan²θ and cosec²θ ≡ 1 + cot²θ (b) the expansions of sin(A ± B), cos(A ± B) and tan(A ± B) (c) the formulae for sin 2A, cos 2A and tan 2A (d) the expression of a sin θ + b cos θ in the forms R sin(θ ± α) and R cos(θ ± α)

Accuracy for this chapter. Exact values stay exact: \(2 - \sqrt3\), \(\tfrac{\sqrt6 + \sqrt2}{4}\), \(-\tfrac{56}{65}\), \(\sqrt{13}\), \(\tfrac{\pi}{8}\). Angles in degrees to 1 decimal place; radians to 3 significant figures unless exact. Write \(\alpha\) to 2 decimal places (\(33.69^\circ\)) but keep the unrounded value (33.6901°) in the calculator for the next part. Write the equation you are solving (\(\tan\theta = 1.5\), \(\cos(\theta - \alpha) = \tfrac{1}{\sqrt{13}}\)) before every angle: an angle with no method line earns nothing.

• “\(\cos(A + B) = \cos A\cos B + \sin A\sin B\).” Correct \(\cos(A + B) \equiv \cos A\cos B - \sin A\sin B\): cos(A + B) has a minus sign. MF19 prints \(\cos(A \pm B) \equiv \cos A\cos B \mp \sin A\sin B\); the \(\mp\) means the sign on the right is the opposite of the sign on the left. In the sine formula the signs match; in the tangent formula the numerator matches and the denominator is opposite.
• “\(\sin 2\theta = \sin\theta\), so divide by \(\sin\theta\): \(2\cos\theta = 1\).” Correct Never divide an equation by \(\sin\theta\) or \(\cos\theta\): the solutions where it is zero vanish. Collect on one side and factorise: \(\sin\theta(2\cos\theta - 1) = 0\), so \(\sin\theta = 0\) or \(\cos\theta = \tfrac12\). On \(0^\circ \le \theta \le 360^\circ\) that is five solutions, and dividing keeps only two.
• “\(\sec\theta = \dfrac{1}{\sin\theta}\).” Correct Sec goes with cos, cosec with sin: \(\sec\theta = \dfrac{1}{\cos\theta}\), \(\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}\). Each pair has exactly one “co”. And \(\sec x\) is not \(\cos^{-1}x\): one is a reciprocal, the other an inverse function.
• “\(\sec\theta = \tfrac12\), so \(\theta = \cos^{-1} 2\) … calculator error.” Correct \(|\sec\theta| \ge 1\) always, so \(\sec\theta = \tfrac12\) has no solutions. Reject it in writing, with the reason: “\(\cos\theta = 2\) is impossible since \(-1 \le \cos\theta \le 1\).”
• “\(\sqrt{13}\cos(\theta - 33.69^\circ) = 1\): \(\theta - 33.69^\circ = 73.90^\circ\), so \(\theta = 107.6^\circ\). Done.” Correct Shift the interval first. If \(0^\circ < \theta < 360^\circ\) then \(-33.69^\circ < \theta - \alpha < 326.31^\circ\). Every value of the bracket in that interval is a solution: \(73.90^\circ\) and \(286.10^\circ\), so \(\theta = 107.6^\circ\) and \(319.8^\circ\).
• “The greatest value of \(3\cos\theta + 2\sin\theta\) is \(3 + 2 = 5\).” Correct The two terms do not peak at the same \(\theta\). Written as \(\sqrt{13}\cos(\theta - \alpha)\), the greatest value is \(R = \sqrt{13} \approx 3.61\).
• “\(\sec\theta = \dfrac{1}{\sin\theta}\).” Repair Sec goes with cos: \(\sec\theta = \dfrac{1}{\cos\theta}\); \(\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}\).
• “\(\sec x = \cos^{-1}x\).” Repair \(\cos^{-1}x\) is the inverse function (the angle whose cosine is \(x\)); \(\sec x = (\cos x)^{-1} = \dfrac{1}{\cos x}\) is a reciprocal. Different objects.
• “\(\sec\theta = \tfrac12 \Rightarrow \theta = \cos^{-1}2\).” Repair \(\cos\theta = 2\) is impossible; \(\sec\theta\) never lies in \((-1, 1)\), so there are no solutions. Say so in words.
• “\(\sec^2\theta = 1 - \tan^2\theta\).” Repair Divide \(\sin^2\theta + \cos^2\theta \equiv 1\) by \(\cos^2\theta\): \(\tan^2\theta + 1 \equiv \sec^2\theta\). The booklet prints it; copy it.
• “\(\sin(A + B) = \sin A + \sin B\).” Repair \(\sin(A + B) \equiv \sin A\cos B + \cos A\sin B\). Test with \(A = B = 30^\circ\): \(\sin 60^\circ = 0.866\), but \(2\sin 30^\circ = 1\).
• “\(\cos(A + B) = \cos A\cos B + \sin A\sin B\).” Repair The sign on the right is opposite to the sign on the left (\(\mp\)): \(\cos(A + B) \equiv \cos A\cos B - \sin A\sin B\).
• “\(\tan(A - B) = \dfrac{\tan A - \tan B}{1 - \tan A\tan B}\).” Repair The denominator carries the opposite sign: \(1 + \tan A\tan B\).
• “\(\cos 2\theta = 2\cos\theta\)” or “\(\sin 2\theta = 2\sin\theta\).” Repair \(\sin 2\theta \equiv 2\sin\theta\cos\theta\); \(\cos 2\theta\) has three forms, none of them \(2\cos\theta\).
• “\(\sin 2\theta = \sin\theta\): divide by \(\sin\theta\), \(\cos\theta = \tfrac12\).” Repair Factorise: \(\sin\theta(2\cos\theta - 1) = 0\). \(\sin\theta = 0\) gives three more solutions on \(0^\circ \le \theta \le 360^\circ\).
• “\(\cos 2\theta + 3\sin\theta = 2\): use \(\cos 2\theta = 2\cos^2\theta - 1\).” Repair Choose the form that leaves one function: with \(\sin\theta\) present, use \(1 - 2\sin^2\theta\).
• “\(R = a + b\)” or “\(R = a^2 + b^2\).” Repair \(R = \sqrt{a^2 + b^2}\), from \(R^2\cos^2\alpha + R^2\sin^2\alpha = a^2 + b^2\).
• “\(\tan\alpha = \dfrac{a}{b}\) for \(a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)\).” Repair Expand: \(R\cos\alpha = a\), \(R\sin\alpha = b\), so \(\tan\alpha = \dfrac{b}{a}\). Match; do not guess.
• “\(\cos(\theta - 33.69^\circ) = 0.277\), so \(\theta - \alpha = 73.90^\circ\). One answer.” Repair \(\theta - \alpha\) ranges over \((-33.69^\circ, 326.31^\circ)\); find every value of the bracket in that interval (\(73.90^\circ\) and \(286.10^\circ\)), then add \(\alpha\).
• “The greatest value of \(3\cos\theta + 2\sin\theta\) is 5.” Repair The two terms do not peak at the same \(\theta\); the greatest value is \(R = \sqrt{13}\).

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. Chapter 12 differentiates \(\sin x\), \(\cos x\) and \(\tan x\), and needs radians to do it: re-answer drill item 6 of card E (in radians) when you get there. Chapter 13 integrates \(\sin^2 x\) and \(\cos^2 x\) through \(\sin^2 x \equiv \tfrac12(1 - \cos 2x)\): re-answer retrieval question 14. Chapter 17 writes complex numbers in the polar form \(r(\cos\theta + i\sin\theta)\), and multiplying two of them uses the compound-angle formulae: re-answer retrieval question 10. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: chapters 12, 13 and 17 use these identities without re-teaching them, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the subject content for that paper, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 3 content, examined in Paper 3 (and Paper 2 on the AS-only pure route). Paper 3 (Pure Mathematics 3) is compulsory for the A Level and is 30% of it. The Paper 2 (Pure Mathematics 2) outcomes this chapter also serves are marked in the syllabus map: Paper 2 is offered only in the AS-only Pure Mathematics route (Papers 1 and 2), where it is 40% of the AS Level, and that route cannot be carried forward to the A Level. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A show that an identity holds, followed by hence solve an equation that the identity simplifies; an express in the form \(R\cos(\theta - \alpha)\) followed by an equation and a greatest or least value; an equation in an interval, in degrees or radians, where choosing the identity is the first step. In each, the reasoning sits in the first line of working — the identity substituted — and in the last, where every solution in the interval is listed and every impossible value rejected with its reason.
• MF19 prints \(\tan\theta \equiv \sin\theta/\cos\theta\), the three Pythagorean identities, all three compound-angle formulae, \(\sin 2A\), \(\cos 2A\) in its three forms and \(\tan 2A\). It does not print the definitions of sec, cosec and cot, the exact values, the graphs, \(\sin^2 A \equiv \tfrac12(1 - \cos 2A)\) or \(\cos^2 A \equiv \tfrac12(1 + \cos 2A)\), or anything about the R-form: those must be known. The MF19 card lists both sides.
• Exact values stay exact (\(2 - \sqrt3\), \(\tfrac{56}{65}\), \(\tfrac{\pi}{8}\)). Angles in degrees to 1 decimal place, radians to 3 significant figures unless exact. \(\alpha\) in the R-form is written to 2 decimal places and carried unrounded into the equation, or the final angle can drift by 0.1°. The equation solved for a principal value must be written before the angle, and every further solution justified from the interval.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.