HookHow Britain was mapped with triangles
Between 1935 and 1962 the Ordnance Survey re-mapped the whole of Great Britain by covering it in triangles. Surveyors built around 6,500 concrete trig pillars on hilltops — the first at Cold Ashby in Northamptonshire on 18 April 1936 — sited so that from any one pillar you could see at least two others. Measure a single baseline on the ground with great care, then measure only the angles between distant pillars, and trigonometry converts those angles into every distance you never walked. The sine and cosine rules turned a handful of sightings into the coordinates of an entire nation; the same pillars still stand on British hills today.
That is the reach of this chapter. Trigonometry begins as the ratios in a right-angled triangle but becomes the mathematics of anything that rotates, oscillates or repeats — tides, alternating current, sound, a crane's jib, a projectile's flight. You will extend sine, cosine and tangent to every angle using the unit circle (5.1, 5.3), solve triangles that are not right-angled and measure angles in radians (5.1), meet the reciprocal and inverse functions (5.4), wield the identities that rewrite one expression as another (5.5, 5.6), solve equations whose unknown is an angle (5.7), prove results rigorously (5.8), and apply the whole toolkit to vectors, forces and motion (5.9) — plus two small-angle approximations (5.2) that quietly power the physics of pendulums and lenses.
ModelFrom triangle to circle — definitions, graphs and the exact values
In a right-angled triangle the ratios are SOH-CAH-TOA: \(\sin\theta=\tfrac{\text{opp}}{\text{hyp}}\), \(\cos\theta=\tfrac{\text{adj}}{\text{hyp}}\), \(\tan\theta=\tfrac{\text{opp}}{\text{adj}}\). That only reaches angles between \(0\) and \(90^\circ\). The unit circle extends the definitions to every angle: for a point that has rotated through \(\theta\) on a circle of radius 1, \(\cos\theta\) is its \(x\)-coordinate, \(\sin\theta\) its \(y\)-coordinate, and \(\tan\theta=\tfrac{\sin\theta}{\cos\theta}\) the gradient of the radius. The signs then follow the quadrant (the CAST diagram): all positive in the first, only sine in the second, only tangent in the third, only cosine in the fourth.
Graphing these, \(y=\sin\theta\) and \(y=\cos\theta\) are waves of amplitude 1 and period \(360^\circ\ (2\pi)\), the cosine curve being the sine curve shifted left by \(90^\circ\). The tangent graph has period \(180^\circ\) with vertical asymptotes wherever \(\cos\theta=0\). Their symmetries are exam gold: \(\sin(-\theta)=-\sin\theta\) (odd), \(\cos(-\theta)=\cos\theta\) (even), and \(\sin(180^\circ-\theta)=\sin\theta\). Finally, the exact values come from two special triangles — the \(45\!-\!45\!-\!90\) and the \(30\!-\!60\!-\!90\) — giving \(\sin30^\circ=\tfrac12\), \(\cos30^\circ=\tfrac{\sqrt3}{2}\), \(\tan30^\circ=\tfrac{1}{\sqrt3}\), \(\sin45^\circ=\cos45^\circ=\tfrac{1}{\sqrt2}\), \(\sin60^\circ=\tfrac{\sqrt3}{2}\), and \(\tan60^\circ=\sqrt3\). Learn them as surds — a non-calculator paper expects them.
MechanismTriangles without a right angle — the sine and cosine rules, and radians
When a triangle has no right angle, two rules do the work. The sine rule \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\) links a side to the angle opposite it — use it when you have a matching side-and-angle pair. The cosine rule \(a^{2}=b^{2}+c^{2}-2bc\cos A\) is for two sides and the included angle (to find the third side) or all three sides (to find an angle). The area of any triangle is \(\tfrac12 ab\sin C\), two sides times the sine of the angle between them. Beware the sine rule's ambiguous case: when finding an angle, \(\sin\) also has an obtuse solution \(180^\circ-\theta\), and both may fit.
Radians are the natural unit for the rest of A-Level: one radian is the angle that subtends an arc equal to the radius, so \(2\pi\) radians \(=360^\circ\). In radians the formulae for a sector become clean: arc length \(s=r\theta\) and sector area \(A=\tfrac12 r^{2}\theta\). Radians are compulsory for calculus of trig functions and for the small-angle results below, because those all rely on the fact that, in radians, \(\sin\theta\approx\theta\) for small \(\theta\).
The triangulation method itself. Two trig pillars \(A\) and \(B\) sit on a carefully measured baseline \(5.0\) km apart. From \(A\) the angle to a distant church spire \(C\) is \(40^\circ\); from \(B\) the angle to the same spire is \(65^\circ\). Find the distance \(AC\) without going near the spire.
The angle at the spire is \(C=180^\circ-40^\circ-65^\circ=75^\circ\). Apply the sine rule with \(AC\) opposite the \(65^\circ\) angle at \(B\), and the baseline \(AB\) opposite the \(75^\circ\) angle at \(C\): \[AC=\dfrac{AB\,\sin B}{\sin C}=\dfrac{5.0\times\sin65^\circ}{\sin75^\circ}=\dfrac{5.0\times0.9063}{0.9659}=4.69\text{ km}.\] From one measured baseline and two angles, an unreachable distance drops out — exactly how the Ordnance Survey fixed the position of every landmark it never physically reached.
MechanismSmall-angle approximations — why a pendulum obeys a neat formula
For a small angle \(\theta\) measured in radians, three approximations hold: \[\sin\theta\approx\theta,\qquad \tan\theta\approx\theta,\qquad \cos\theta\approx1-\tfrac{\theta^{2}}{2}.\] They come from truncating the functions' power series, and geometrically from the fact that for a thin wedge of the unit circle the arc, the chord and the tangent are almost the same length. The radian requirement is absolute — the results are false in degrees, because they encode the calculus fact that \(\dfrac{\sin\theta}{\theta}\to1\) only when \(\theta\) is in radians.
These are not just curiosities. The entire simple-harmonic theory of a pendulum uses \(\sin\theta\approx\theta\) to replace an unsolvable equation with the tidy \(T=2\pi\sqrt{L/g}\); lens and mirror optics (the paraxial approximation) rest on the same step. In exam terms, they turn otherwise-impossible limits into instant simplifications.
Estimate \(\dfrac{1-\cos2\theta}{\theta\,\sin\theta}\) for small \(\theta\), a classic exam use of the approximations.
Replace each piece with its small-angle form. The numerator: \(\cos2\theta\approx1-\tfrac{(2\theta)^{2}}{2}=1-2\theta^{2}\), so \(1-\cos2\theta\approx2\theta^{2}\). The denominator: \(\sin\theta\approx\theta\), so \(\theta\sin\theta\approx\theta^{2}\). Therefore \[\dfrac{1-\cos2\theta}{\theta\,\sin\theta}\approx\dfrac{2\theta^{2}}{\theta^{2}}=2.\] A direct numerical check with \(\theta=0.05\) rad gives \(0.0049958/0.0024998=1.9985\), converging on 2 as \(\theta\to0\). The powers of \(\theta\) had to match top and bottom for a finite answer — if they had not, you would have taken more terms of the \(\cos\) expansion.
ModelThe reciprocal and inverse functions, and the three Pythagorean identities
Three reciprocal functions complete the family: \(\sec\theta=\dfrac{1}{\cos\theta}\), \(\operatorname{cosec}\theta=\dfrac{1}{\sin\theta}\), and \(\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{\cos\theta}{\sin\theta}\). Each has vertical asymptotes wherever its parent is zero, so \(\sec\) and \(\tan\) 'blow up' at the same angles. The inverse functions \(\arcsin\), \(\arccos\) and \(\arctan\) undo sine, cosine and tangent, but only after the domains are restricted so each is one-to-one: \(\arcsin\) returns angles in \([-90^\circ,90^\circ]\), \(\arccos\) in \([0^\circ,180^\circ]\), and \(\arctan\) in \((-90^\circ,90^\circ)\). That restriction is exactly why your calculator gives only one 'principal value' when you solve an equation.
From the unit circle, \(x^{2}+y^{2}=1\) gives the master identity \[\sin^{2}\theta+\cos^{2}\theta=1.\] Divide it through by \(\cos^{2}\theta\) and you get \(\tan^{2}\theta+1=\sec^{2}\theta\); divide instead by \(\sin^{2}\theta\) and you get \(1+\cot^{2}\theta=\operatorname{cosec}^{2}\theta\). Together with \(\tan\theta=\tfrac{\sin\theta}{\cos\theta}\), these are the levers for nearly every simplification and proof in the chapter, and they reappear as the key substitutions in integration.
Given \(\sin\theta=\tfrac{3}{5}\) with \(\theta\) acute, find the other five ratios without a calculator. From \(\sin^{2}\theta+\cos^{2}\theta=1\), \(\cos^{2}\theta=1-\tfrac{9}{25}=\tfrac{16}{25}\), and since \(\theta\) is acute \(\cos\theta=\tfrac{4}{5}\). Then \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\tfrac{3}{4}\), and the reciprocals fall out: \(\operatorname{cosec}\theta=\tfrac{5}{3}\), \(\sec\theta=\tfrac{5}{4}\), \(\cot\theta=\tfrac{4}{3}\). Recognising the 3-4-5 triangle is faster than any button, and keeping the answers as exact fractions is what a non-calculator paper wants.
MechanismCompound and double angles, and folding a wave into one sinusoid
The compound-angle formulae handle the sine and cosine of a sum: \[\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B,\qquad \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B,\] with \(\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}\). Putting \(B=A\) gives the double-angle formulae: \(\sin2A=2\sin A\cos A\), and \(\cos2A=\cos^{2}A-\sin^{2}A=2\cos^{2}A-1=1-2\sin^{2}A\) (three interchangeable forms — pick the one that cancels what you already have). These are the engine of trig proof and of integrating \(\cos^{2}\) and \(\sin^{2}\).
The harmonic (R) form folds a sum of a sine and a cosine of the same angle into a single shifted wave: \[a\cos\theta+b\sin\theta=R\cos(\theta-\alpha),\quad R=\sqrt{a^{2}+b^{2}},\ \tan\alpha=\tfrac{b}{a}.\] Because a single sinusoid has an obvious maximum and minimum \((\pm R)\), the R-form is the standard route to the greatest value of a combined oscillation and to solving equations like \(a\cos\theta+b\sin\theta=c\).
Express \(3\sin\theta+4\cos\theta\) in the form \(R\sin(\theta+\alpha)\), then state its maximum value and where it occurs.
Expand the target: \(R\sin(\theta+\alpha)=R\cos\alpha\,\sin\theta+R\sin\alpha\,\cos\theta\). Matching coefficients, \(R\cos\alpha=3\) and \(R\sin\alpha=4\). Squaring and adding, \(R^{2}=3^{2}+4^{2}=25\), so \(R=5\); dividing, \(\tan\alpha=\tfrac{4}{3}\), so \(\alpha=53.13^\circ\). Hence \[3\sin\theta+4\cos\theta=5\sin(\theta+53.13^\circ).\] The maximum of \(\sin\) is 1, so the greatest value is \(5\), reached when \(\theta+53.13^\circ=90^\circ\), i.e. \(\theta=36.87^\circ\). Two coefficients have become one amplitude and one phase shift — the same trick engineers use to combine two out-of-step AC signals into one.
MechanismSolving trigonometric equations across an interval
A trig equation almost always has several solutions in the given range, and the marks are for finding all of them. The method: reduce the equation to the form \(\sin X=k\), \(\cos X=k\) or \(\tan X=k\); take the calculator's principal value; then use the graph's symmetry and periodicity to generate the rest inside the interval. For a quadratic in a trig function, substitute (let \(s=\sin\theta\)), factorise, solve each root separately, and discard any root with \(|s|>1\) as impossible.
Two traps dominate the mark schemes. First, when the argument is a multiple angle such as \(2\theta\) or \(3\theta\), widen the search interval to match before solving — for \(0\le\theta<360^\circ\) you must find every \(2\theta\) in \(0\le2\theta<720^\circ\), then halve. Second, never divide an equation by \(\cos\theta\) (or \(\sin\theta\)) to simplify it: that silently deletes the solutions where \(\cos\theta=0\). Factorise instead.
Solve \(2\sin^{2}\theta-\sin\theta-1=0\) for \(0^\circ\le\theta<360^\circ\).
Let \(s=\sin\theta\). The quadratic \(2s^{2}-s-1=0\) factorises as \((2s+1)(s-1)=0\), giving \(s=-\tfrac12\) or \(s=1\). Take each in turn. \(\sin\theta=1\) has the single solution \(\theta=90^\circ\). \(\sin\theta=-\tfrac12\) has principal value \(-30^\circ\), which is outside the range; using the symmetry of the sine curve, the solutions in \([0^\circ,360^\circ)\) are \(\theta=180^\circ+30^\circ=210^\circ\) and \(\theta=360^\circ-30^\circ=330^\circ\). The full solution set is \(\theta=90^\circ,\,210^\circ,\,330^\circ\). Missing the second or third value — treating the calculator's one answer as the whole story — is the most frequent lost mark on this topic.
CaseProving identities, and trigonometry in forces, motion and vectors
A trig proof (an 'show that' or '\(\equiv\)' question) is a one-way journey: start from one side — usually the more complicated — and manipulate it into the other. Never move terms across the identity sign as if solving. The reliable strategies are to rewrite everything in terms of \(\sin\) and \(\cos\), to deploy \(\sin^{2}\theta+\cos^{2}\theta=1\) (and its two derived forms), and to combine fractions over a common denominator so a Pythagorean identity can collapse them.
The payoff is in context (5.9). Resolving a weight on a slope splits \(mg\) into a component \(mg\sin\theta\) down the incline and \(mg\cos\theta\) pressing into it. A projectile launched at speed \(u\) and angle \(\theta\) has range \(\dfrac{u^{2}\sin2\theta}{g}\), which the double-angle identity shows is greatest at \(\theta=45^\circ\) because \(\sin2\theta\) peaks when \(2\theta=90^\circ\). Bearings, the angle between two vectors, and simple harmonic motion all reduce to the same sine-and-cosine bookkeeping.
Prove that \(\dfrac{1}{1-\sin\theta}+\dfrac{1}{1+\sin\theta}\equiv2\sec^{2}\theta\).
Start from the left and combine over a common denominator: \[\dfrac{(1+\sin\theta)+(1-\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}=\dfrac{2}{1-\sin^{2}\theta}.\] Now the Pythagorean identity replaces the denominator: \(1-\sin^{2}\theta=\cos^{2}\theta\), so the expression becomes \(\dfrac{2}{\cos^{2}\theta}=2\sec^{2}\theta\), which is the right-hand side. Notice the whole proof flowed one way, and the decisive move was recognising \((1-\sin\theta)(1+\sin\theta)\) as a difference of two squares waiting for \(\sin^{2}\theta+\cos^{2}\theta=1\).
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Answer in the form the question demands. On a non-calculator paper, leave exact values as surds — \(\tfrac{\sqrt3}{2}\), not \(0.866\) — because the mark is for the exact form. State which rule you are using and why: sine rule for a side-angle pair, cosine rule for SAS or SSS, and watch the sine rule's obtuse alternative when finding an angle.
When solving over an interval, the discipline that earns full marks is finding every solution: sketch the relevant curve, mark the principal value, and read off the symmetric partners; if the argument is \(2\theta\), widen the interval to \(0\le2\theta<720^\circ\) first, solve, then divide. For the R-form, quote \(R\) exactly and \(\alpha\) to a sensible accuracy, and read the maximum straight off as \(\pm R\). In proofs, commit to one side and transform it — a proof that shuttles terms across the \(\equiv\) sign scores nothing. And switch to radians the moment calculus or a small-angle approximation appears, since both are only valid there.