Trigonometry

📚The mathematics

Six functions, any angle

For angles beyond \( 90^{\circ} \), the sign depends on the quadrant: all positive in the first, sine in the second, tangent in the third, cosine in the fourth (“CAST” going anticlockwise from the fourth). The size comes from the related acute angle. So \( \cos 240^{\circ} = -\cos 60^{\circ} = -\tfrac{1}{2} \), and \( \sec 240^{\circ} = -2 \).

Paper 1 expects the exact values:

Amplitude, period and the graphs

For \( y = a\sin bx + c \) and \( y = a\cos bx + c \) (with \( a > 0 \)):

For \( y = a\tan bx + c \) the period is \( \dfrac{180^{\circ}}{b} \) (or \( \dfrac{\pi}{b} \)), there is no amplitude, and there are vertical asymptotes where \( bx = 90^{\circ}, 270^{\circ}, \ldots \). The syllabus says the \(x\)-coordinate of each asymptote must be labelled.

The identities

All three are in the List of formulas; the second and third are the first divided by \( \cos^{2}A \) and by \( \sin^{2}A \). Use them to turn an equation into a single function (usually a quadratic in it), and to prove identities.

Solving equations in a domain

1. Rearrange to one function equal to a number (use an identity, or divide \( \sin \) by \( \cos \) to get \( \tan \)).
2. If the angle is compound, such as \( 2x + \tfrac{\pi}{6} \), transform the domain first: if \( 0 \le x \le \pi \) then \( \tfrac{\pi}{6} \le 2x + \tfrac{\pi}{6} \le 2\pi + \tfrac{\pi}{6} \).
3. Find the principal value, then every other value in the (transformed) domain, using the symmetry of the graph.
4. Undo the compound angle, and check every answer lies in the original domain.

Never cancel a factor such as \( \sin x \) from both sides — factorise instead, or you lose the solutions of \( \sin x = 0 \). The syllabus also warns you that the domain may be in degrees or radians: answer in the same unit, to 1 d.p. for degrees and 3 s.f. for radians unless told otherwise.

Proving identities

Start from the more complicated side and work towards the other; never work on both sides at once. Writing everything in terms of \( \sin \) and \( \cos \), and combining fractions over a common denominator, solves most of them. Finish by stating the other side.

✏️Worked example

(a) Solve \( 3\sec^{2}\theta - 5\tan\theta - 1 = 0 \) for \( 0^{\circ} \le \theta \le 360^{\circ} \). [5] (b) Solve \( 2\cos\left(2x + \tfrac{\pi}{6}\right) = \sqrt{3} \) for \( 0 \le x \le \pi \), giving exact answers. [4] (c) Prove that \( (1 - \sin x)(\sec x + \tan x) = \cos x \). [3]

(a) Replace \( \sec^{2}\theta \) by \( 1 + \tan^{2}\theta \): \( 3 + 3\tan^{2}\theta - 5\tan\theta - 1 = 0 \), so \( 3\tan^{2}\theta - 5\tan\theta + 2 = 0 \) and \( (3\tan\theta - 2)(\tan\theta - 1) = 0 \).

• \( \tan\theta = 1 \): \( \theta = 45^{\circ} \) or \( 45^{\circ} + 180^{\circ} = 225^{\circ} \).
• \( \tan\theta = \tfrac{2}{3} \): \( \theta = 33.7^{\circ} \) or \( 213.7^{\circ} \).

(b) \( \cos\left(2x + \tfrac{\pi}{6}\right) = \tfrac{\sqrt{3}}{2} \). With \( 0 \le x \le \pi \), the compound angle runs over \( \tfrac{\pi}{6} \le 2x + \tfrac{\pi}{6} \le \tfrac{13\pi}{6} \). In that interval \( \cos u = \tfrac{\sqrt{3}}{2} \) at \( u = \tfrac{\pi}{6},\ \tfrac{11\pi}{6},\ \tfrac{13\pi}{6} \). So \( 2x = 0,\ \tfrac{5\pi}{3},\ 2\pi \), giving

\[ x = 0,\quad \frac{5\pi}{6},\quad \pi \]

(c) Start from the left-hand side and write the bracket in terms of \( \sin \) and \( \cos \):

\[ (1 - \sin x)\left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right) = \frac{(1 - \sin x)(1 + \sin x)}{\cos x} = \frac{1 - \sin^{2}x}{\cos x} = \frac{\cos^{2}x}{\cos x} = \cos x \]

Check it. (a) \( \tan 33.7^{\circ} = 0.667 \) and \( 3\sec^{2}33.7^{\circ} = 4.33 \): \( 4.33 - 3.33 - 1 = 0 \) ✓. (b) At \( x = \tfrac{5\pi}{6} \), \( 2x + \tfrac{\pi}{6} = \tfrac{11\pi}{6} \) and \( 2\cos\tfrac{11\pi}{6} = \sqrt{3} \) ✓.

Forgetting to stretch the domain. In (b), solving \( \cos u = \tfrac{\sqrt{3}}{2} \) only on \( 0 \le u \le 2\pi \) finds \( \tfrac{\pi}{6} \) and \( \tfrac{11\pi}{6} \) and misses \( \tfrac{13\pi}{6} \) — and with it the solution \( x = \pi \). The doubled angle goes round the circle more than once.