Geometry and trigonometry

📚The mathematics

3.1 Three dimensions

The distance between two points in space extends Pythagoras by one term: \( d = \sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2} + (z_2-z_1)^{2}} \), and the midpoint is the average of each coordinate. Volumes and surface areas of the pyramid, cone, sphere, hemisphere and their combinations are in the formula booklet.

The real skill in three-dimensional questions is not the formulae but finding the right-angled triangle. At standard level every 3D trigonometry question reduces to a right-angled triangle you have to spot inside the solid. Draw that triangle separately, in two dimensions, with its lengths marked. Trying to do trigonometry on the perspective drawing itself is how people talk themselves into using the wrong side.

3.2 & 3.3 Solving triangles

In a right-angled triangle use \( \sin, \cos, \tan \) directly. In any other triangle:

Choose by counting what you have. The sine rule needs a matched pair — a side and the angle opposite it. If you have no such pair, you need the cosine rule: use the first form when you know two sides and the angle between them, and the rearranged form when you know all three sides and want an angle. The area formula needs two sides and the included angle, the one between them.

Angles of elevation and depression are measured from the horizontal, and the angle of depression from A to B equals the angle of elevation from B to A. Bearings are three figures, measured clockwise from north: due east is 090°, south-west is 225°. Always draw the north line at each point you take a bearing from.

3.4 Radians

One radian is the angle subtending an arc equal to the radius, so \( \pi \) radians is 180°. Multiply by \( \dfrac{\pi}{180} \) to convert degrees to radians and by \( \dfrac{180}{\pi} \) to go back. In radians the arc and sector formulae become as simple as they ever get:

3.5 & 3.6 The unit circle and identities

On a unit circle, the point at angle \( \theta \) has coordinates \( (\cos\theta, \sin\theta) \). Everything else follows. \( \tan\theta = \dfrac{\sin\theta}{\cos\theta} \) is the gradient of the line from the origin, which is why \( \tan \) is undefined where \( \cos\theta = 0 \). The Pythagorean identity is just Pythagoras applied to that point:

The circle also gives you the quadrant relationships without memorising a mnemonic: \( \cos(-x) = \cos x \), \( \sin(-x) = -\sin x \), \( \sin(\pi + x) = -\sin x \), and so on. Sketch the circle, mark the angle, read off the sign.

You are expected to know the exact values for \( 0, \dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}, \dfrac{\pi}{2} \) and their multiples — these are not in the formula booklet and Paper 1 assumes them. The double angle identities are:

Three forms of \( \cos 2\theta \) exist so that you can choose the one containing only the ratio you already know. If you are given \( \sin\theta \), use the last one and no Pythagoras is needed.

3.7 & 3.8 Circular functions and trigonometric equations

For \( f(x) = a\sin\!\left(b(x+c)\right) + d \): \( |a| \) is the amplitude, \( d \) is the principal axis \( y = d \), the period is \( \dfrac{2\pi}{b} \) in radians (or \( \dfrac{360^\circ}{b} \) in degrees), and \( c \) is a horizontal phase shift. Read them off a real graph as \( a = \dfrac{\text{max} - \text{min}}{2} \) and \( d = \dfrac{\text{max} + \text{min}}{2} \). Contexts that produce these models include tides, Ferris wheels, daylight hours and temperature cycles.

To solve a trigonometric equation over an interval, find the first solution, then use the symmetry of the graph or the unit circle to generate every other solution inside the interval. The calculator gives you one value; the marks are for the rest.

AHL 3.9–3.11 Further trigonometry

The reciprocal ratios are \( \sec\theta = \dfrac{1}{\cos\theta} \), \( \csc\theta = \dfrac{1}{\sin\theta} \) and \( \cot\theta = \dfrac{1}{\tan\theta} \). Note the mismatch that catches everyone: secant pairs with cosine. Dividing the Pythagorean identity by \( \cos^{2}\theta \) and then by \( \sin^{2}\theta \) gives the other two:

The inverse functions \( \arcsin \), \( \arccos \) and \( \arctan \) need restricted domains to exist at all, and you should know their ranges — that restriction is exactly why your calculator returns only one solution to a trigonometric equation.

The compound angle identities are the AHL workhorses:

Watch the signs in the cosine and tangent identities: they flip. Setting \( B = A \) in each reproduces the double angle results, including \( \tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^{2}\theta} \) — which is worth doing once so you never have to memorise it separately.

AHL 3.12–3.16 Vectors

A vector has magnitude and direction. In components, \( \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} \), with \( |\mathbf{v}| = \sqrt{v_1^{2}+v_2^{2}+v_3^{2}} \) and unit vector \( \hat{\mathbf{v}} = \dfrac{\mathbf{v}}{|\mathbf{v}|} \). The displacement vector between two points is \( \overrightarrow{AB} = \mathbf{b} - \mathbf{a} \) — destination minus start, and getting that backwards reverses your answer.

The scalar (dot) product returns a number:

so it is the tool for angles, and \( \mathbf{v}\cdot\mathbf{w} = 0 \) is the perpendicularity test. The vector (cross) product returns a vector perpendicular to both, with magnitude \( |\mathbf{v}\times\mathbf{w}| = |\mathbf{v}||\mathbf{w}|\sin\theta \), which is the area of the parallelogram they span — so half of it is the area of the triangle. It is zero exactly when the vectors are parallel.

AHL 3.14–3.18 Lines and planes

A line is \( \mathbf{r} = \mathbf{a} + \lambda\mathbf{b} \): a point you know plus a direction you travel in. The same line has infinitely many valid equations, since any point on it and any multiple of the direction will do — so an answer that differs from the markscheme is not automatically wrong. In kinematics \( \lambda \) is time and \( |\mathbf{b}| \) is the speed.

Two lines in space are coincident, parallel, intersecting, or skew — non-parallel and non-intersecting, which has no two-dimensional analogue. Test by first comparing directions for parallelism, then solving two of the three component equations simultaneously and checking the third. If the third fails, the lines are skew. That check is the whole question.

A plane is \( \mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n} \), with \( \mathbf{n} \) the normal, or in Cartesian form \( ax + by + cz = d \) where \( (a,b,c) \) is the normal vector. Angles involving planes are computed through the normal, so remember to take the complement when you want the angle between a line and a plane rather than between the line and the normal. Intersections of three planes are the geometry behind the systems of equations in Topic 1: a point, a line, or nothing at all.