Trigonometry

• The calculator gives one angle. The graph gives all of them: second value from the symmetry, the rest by the period, and nothing outside the interval.
• Interval for \(X\) first. Second value: \(180^\circ - \alpha\) for sin, \(360^\circ - \alpha\) for cos. Add the period, discard, convert back, count.

• 1.5.1 I can sketch — sketch and use graphs of the sine, cosine and tangent functions (for angles of any size, and using either degrees or radians)
• 1.5.2 I can use — use the exact values of the sine, cosine and tangent of 30°, 45°, 60°, and related angles
• 1.5.3 I can use — use the notations sin⁻¹x, cos⁻¹x, tan⁻¹x to denote the principal values of the inverse trigonometric relations
• 1.5.4 I can use — use the identities sin θ / cos θ ≡ tan θ and sin²θ + cos²θ ≡ 1
• 1.5.5 I can find — find all the solutions of simple trigonometrical equations lying in a specified interval (general forms of solution are not included)

Why the habit has to go. In a triangle an angle is between 0° and 180°, so “the angle whose sine is \(\tfrac12\)” could only be 30° or 150°, and the diagram told you which. On a graph that runs forever, \(\sin x = \tfrac12\) has infinitely many solutions. The question now always names an interval, and the answer is every solution in it.

• “\(3\sin 2x + 1 = 0\) for \(-\pi < x < \pi\): \(\sin 2x = -\tfrac13\), so \(2x = -0.340\) or \(3.48\), and \(x = -0.170\) or \(1.74\).” Correct Find the interval for \(2x\) before solving. If \(-\pi < x < \pi\) then \(-2\pi < 2x < 2\pi\): two full turns, so four values of \(2x\), found by adding and subtracting \(2\pi\) before halving. The answer has four solutions, \(x = -1.40, -0.170, 1.74, 2.97\).
• “\(\sin x = 0.4\), so \(x = \sin^{-1}0.4 = 23.6^\circ\).” Correct The principal value is one solution, not all of them. Sine is also positive in the second quadrant, so \(180^\circ - 23.6^\circ = 156.4^\circ\) is a solution too, and a wider interval adds more by the period. The calculator gives the principal value; the graph gives the rest.
• “\(\cos x = -0.3\): the second value is \(180^\circ - 107.5^\circ\).” Correct \(180^\circ - \alpha\) is the second value for sine. The cosine graph is symmetric about \(x = 0\), so its second value is \(-\alpha\), which is \(360^\circ - \alpha = 252.5^\circ\) in \(0^\circ \le x \le 360^\circ\).
• “\(3\sin x\cos x = 2\cos x\), so dividing by \(\cos x\), \(\sin x = \tfrac23\).” Correct Dividing by \(\cos x\) throws away every solution of \(\cos x = 0\). Factorise: \(\cos x(3\sin x - 2) = 0\), and solve both factors.
• “\(y = \tan(x + \tfrac14\pi)\) is \(y = \tan x\) moved \(\tfrac14\pi\) to the right.” Correct \(y = f(x + a)\) is a translation by \(\begin{pmatrix} -a \\ 0 \end{pmatrix}\), to the left. The asymptotes move from \(\tfrac{\pi}{2}, \tfrac{3\pi}{2}\) to \(\tfrac{\pi}{4}, \tfrac{5\pi}{4}\).
• “\(\sin^{-1}x\) means \(\dfrac{1}{\sin x}\).” Correct The \(-1\) is the inverse-function notation of chapter 2. \(\sin^{-1}\tfrac12 = \tfrac{\pi}{6}\), an angle; \(\dfrac{1}{\sin(\frac12)}\) is a number near 2.09.
• “\(3\cos^2\theta + 5\cos\theta - 2 = 0\) gives \(\cos\theta = \tfrac13\) or \(\cos\theta = -2\), so …” Correct \(-1 \le \cos\theta \le 1\) for every \(\theta\), so \(\cos\theta = -2\) has no solutions. Reject it with the reason written, then solve \(\cos\theta = \tfrac13\) fully.
• “The period of \(y = 1 - \cos 2x\) is \(720^\circ\).” Repair \(y = f(2x)\) is a stretch parallel to the \(x\)-axis with factor \(\tfrac12\): it squashes the graph, so the period halves to \(180^\circ\) (\(\pi\)).
• “\(y = \tan(x + \tfrac14\pi)\) is \(y = \tan x\) moved \(\tfrac14\pi\) to the right.” Repair It is the translation by \(\begin{pmatrix} -\frac14\pi \\ 0 \end{pmatrix}\), to the left; asymptotes at \(\tfrac{\pi}{4}\) and \(\tfrac{5\pi}{4}\).
• Drawing \(y = \tan x\) as one continuous curve running through \(x = 90^\circ\). Repair \(\tan 90^\circ\) is undefined. The branches are separate, each approaching the dashed asymptote without touching it.
• “\(\cos 150^\circ = \tfrac12\sqrt3\).” Repair 150° is in the second quadrant, where cos is negative: \(\cos 150^\circ = -\cos 30^\circ = -\tfrac12\sqrt3\).
• “\(\sin^{-1}(-\tfrac12) = 330^\circ\).” Repair The principal value lies in \(-90^\circ\) to \(90^\circ\): \(-30^\circ\). 330° is another solution of \(\sin x = -\tfrac12\), not the principal value.
• “\(\sin^{-1}x = \dfrac{1}{\sin x}\).” Repair \(\sin^{-1}\) is the inverse function; \(\dfrac{1}{\sin x}\) is the reciprocal of \(\sin x\), a different thing altogether.
• “\(\sin^2\theta + \cos^2\theta \equiv 1\), so \(\sin\theta + \cos\theta = 1\).” Repair The identity is about squares. \(\sin 30^\circ + \cos 30^\circ = 0.5 + 0.866 = 1.366\), not 1.
• Proving an identity by writing it down and cross-multiplying both sides. Repair That starts from the result. Start from one side and reach the other, every line following from the one before.
• “\(\sin x = 0.4 \Rightarrow x = 23.6^\circ\) only.” Repair Sine is also positive in the second quadrant: \(180^\circ - 23.6^\circ = 156.4^\circ\) is a solution as well.
• “\(\cos x = -0.3\), second solution \(180^\circ - 107.5^\circ\).” Repair For cos the second value is \(360^\circ - \alpha\): \(360^\circ - 107.5^\circ = 252.5^\circ\). \(72.5^\circ\) has positive cosine and is not a solution.
• “\(3\sin 2x + 1 = 0\): \(\sin 2x = -\tfrac13\), \(2x = -0.340, 3.48\), so \(x = -0.170, 1.74\).” Repair \(2x\) runs over \((-2\pi, 2\pi)\); add and subtract \(2\pi\) before halving. There are four solutions: \(-1.40, -0.170, 1.74, 2.97\).
• Dividing \(3\sin x\cos x = 2\cos x\) by \(\cos x\). Repair Factorise: \(\cos x(3\sin x - 2) = 0\). \(\cos x = 0\) gives \(90^\circ\) and \(270^\circ\), which division loses.
• “\(3\cos^2\theta + 5\cos\theta - 2 = 0\) gives \(\cos\theta = -2\), \(\theta = \ldots\)” Repair \(|\cos\theta| \le 1\); reject \(-2\) and write the reason.
• Giving \(70.5288^\circ\) as the final answer. Repair Degrees to 1 decimal place: \(70.5^\circ\). Radians to 3 significant figures; exact where the question asks for exact.

• 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.
• Two checks that catch most sign errors. A sketch of the graph: \(\cos x\) is below the axis between 90° and 270°, so \(\cos 150^\circ\) must be negative. And a calculator check after the exact work: \(-\tfrac12\sqrt3 \approx -0.866\) and \(\cos 150^\circ\) on the calculator agree. The exact form is still what you write.
• Interleave with the chapters that use this one. Chapter 11 (Paper 3) adds the reciprocal functions and the compound- and double-angle formulae: when you reach it, re-solve worked example 6 and ask which new identity would have produced the same quadratic. Chapter 12 differentiates these functions: redraw figure 1 and say where the gradient of \(\sin x\) is zero. Chapter 18 (Mechanics) resolves forces with sine and cosine: re-answer the exact-value drill in degrees. Chapter 17 writes complex numbers in polar form: re-answer the principal-value drill. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: Paper 1 is assumed knowledge for every other paper, and an individual examination question may involve ideas and methods from more than one section of that paper’s content, 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 1 content, examined in Paper 1. Paper 1 (Pure Mathematics 1) is compulsory for both the AS Level and the A Level: it is 60% of the AS Level and 30% of the A Level, and its content is assumed knowledge for every other paper. 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.
• The five outcomes fit together in a few natural ways: a show that identity (1.5.4) that turns a harder equation into one you can solve (1.5.5); a sketch of a transformed graph (1.5.1) used to solve an equation or count its roots; and a direct solve in an interval, perhaps with a multiple angle or a quadratic in \(\cos\theta\). The mixed challenge practises all three. Whatever the form, write the interval for \(2x\), the second value and the reason for rejecting a root, so every step of your reasoning is visible.
• MF19 prints \(\tan\theta \equiv \sin\theta/\cos\theta\), \(\cos^2\theta + \sin^2\theta \equiv 1\) and the principal-value ranges of \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\). The rearrangements \(\sin^2\theta \equiv 1 - \cos^2\theta\) and \(\cos^2\theta \equiv 1 - \sin^2\theta\), the graphs, the exact values and CAST are not printed and must be known. See the MF19 card.
• Answer in the unit the interval uses: an interval in degrees wants degrees to 1 decimal place, one in radians wants radians to 3 significant figures, or exact (\(\pi/12\)) when the question says so. Keep the principal value unrounded (\(-0.339837\ldots\), \(70.5288\ldots^\circ\)) until the last line. Write the principal value and the equation it comes from; a list of angles with no method line is an unsupported calculator answer.
• 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.