3.2 Geometry and trigonometry - AHL content

One radian is the angle subtended by an arc whose length equals the radius. A full turn is 2π radians, so radians connect angle directly to circle geometry.

Convert degrees using θ(rad)=θ°π/180. In radian measure, arc length is s=rθ and sector area is ½r²θ, with θ in radians.

For r=5 and θ=1.2, the arc length is 6 units and the sector area is 15 square units. Using 1.2° instead would give a completely different scale.

Do not substitute degrees into s=rθ or ½r²θ. Check the angle unit before using a formula.

Solving a trigonometric equation means finding every angle in the stated interval that produces the target value. Periodicity creates repeated solutions.

Use the unit circle or graph to identify the reference angle and the quadrants with the correct sign, then add the function's period. A calculator's principal value is only one solution unless the interval makes it complete.

For sin x=0.5 on 0≤x≤2π, x=π/6 and 5π/6. Stopping at π/6 misses the second intersection of the sine curve with 0.5.

The number of solutions depends on the interval and period. Check endpoints and do not use the cosine rule for a sine equation.

Unit-circle definitions are $\cos heta=x$ and $\sin heta=y$ for the point $(x,y)$ on the unit circle, so $\cos^2 heta+\sin^2 heta=1$ and $\tan heta=\sin heta/\cos heta$ when $\cos heta\ne0$. The sine rule's ambiguous case can give two triangles because $\sin heta=\sin(\pi- heta)$; accept only solutions that satisfy the side lengths and stated interval. Exact trig values are helpful but are not directly assessed.

A 2×2 matrix maps a point vector to its image: x′=Ax. The columns show where the basis vectors go, so the matrix encodes the geometric action.

Apply the matrix to the column vector in the stated order. Determinant magnitude gives area scale; a negative determinant reverses orientation. To recover an original point, solve with A⁻¹ when it exists.

The matrix [[0,−1],[1,0]] sends (1,0) to (0,1) and (0,1) to (−1,0): a 90° anticlockwise rotation. Multiplying by the row vector would describe a different operation.

Do not read rows as images of basis vectors without checking the convention. A zero determinant means the transformation has no inverse.

The syllabus also includes affine maps $\begin{pmatrix}x'\y'\end{pmatrix}=A\begin{pmatrix}x\y\end{pmatrix}+\begin{pmatrix}e\f\end{pmatrix}$, so translation requires the added vector and cannot be represented by a $2\times2$ matrix alone. In a composition, the rightmost transformation acts first; $BAx$ means apply $A$, then $B$. Iterating one or more transformation rules from an initial shape or point can generate a fractal, while $|\det A|$ remains the area scale for the linear part.

A vector has magnitude and direction. Two non-zero vectors are parallel when one is a scalar multiple of the other; the scalar's sign tells whether their directions agree or oppose.

Compare components rather than relying on a sketch. The length is √(a·a), and addition represents successive displacements. In 3D, every component must satisfy the same scalar relationship.

(2,−4,6) is 2(1,−2,3), so the vectors are parallel and point the same way. (−2,4,−6) is a negative multiple and points oppositely.

Equal length does not mean parallel, and a single matching component is not enough. Check all components and exclude the zero vector when using direction tests.

A unit vector in the direction of non-zero $v$ is $\hat v=v/|v|$; rescaling to magnitude $s$ gives $s\hat v$. For example, a particle moving at $7\text{ m s}^{-1}$ in direction $3i+4j$ has unit direction $(3i+4j)/5$ and velocity $(21i+28j)/5\text{ m s}^{-1}$. A resultant is the vector sum, and the zero vector has magnitude zero but no defined direction.

A line through point a with direction vector d is r=a+td, where t is a real parameter. Changing t moves along the line without changing its direction.

Use two points P and Q to form d=Q−P. To test whether X lies on the line, solve the component equations for one common t; inconsistent values mean X is not on it.

Through (1,0,2) and (3,4,2), r=(1,0,2)+t(2,4,0). The point (2,2,2) has t=1/2 in every component, so it lies on the line.

Matching one coordinate is not enough. Check the same parameter across all components and distinguish a line from a finite segment.

A vector kinematics model writes position as r(t)=r₀+vt for constant velocity, or uses the corresponding component equations when acceleration is present.

To test whether two moving objects meet, solve their position vectors for a common time in the allowed interval. A closest approach is different: minimise the squared separation rather than forcing an exact intersection.

If r₁=(0,0)+t(3,1) and r₂=(6,4)+t(−1,−1), solving gives t=2 and position (6,2); the same time in both equations is the evidence of an intersection.

Equal coordinates at different times do not mean collision. State the time domain and distinguish position from velocity.

For variable velocity in two dimensions, $v(t)=r'(t)$ and $r(t)=\int v(t)\,dt+C$, with the initial position fixing $C$; speed is $|v(t)|$. Relative position of $B$ from $A$ is $r_B-r_A$. To find closest approach, minimize $|r_B-r_A|^2$ over the allowed time interval and check endpoints as well as stationary values; projectile and circular motion are special cases of this vector model.

The scalar product a·b=|a||b|cosθ measures alignment and is zero for perpendicular vectors. The vector product a×b is perpendicular to both and has magnitude |a||b|sinθ, the area of the parallelogram they span.

Use a·b for angles, projections and perpendicular tests; use a×b for normals, orientation and areas. Order matters for the cross product: b×a=−(a×b).

For a=(1,0,0) and b=(0,2,0), a·b=0 and a×b=(0,0,2), so the vectors are perpendicular and span area 2.

The cross product is not a scalar and the dot product is not a vector. Check the dimension and interpretation of the requested result.

The scalar component of $a$ in the direction of non-zero $b$ is $a\cdot b/|b|=|a|\cos heta$; the magnitude of the component perpendicular to $b$ in their plane is $|a\times b|/|b|=|a|\sin heta$. Thus $|a\times b|$ is parallelogram area and half of it is triangle area. For line directions, use the acute angle by replacing $\cos heta$ with $|a\cdot b|/(|a||b|)$.

A graph consists of vertices and edges representing objects and connections. Edges may be directed or undirected and weighted or unweighted, depending on the relationship being modelled.

The degree counts incident edges; a path records a route and a cycle returns to its start. Choose the graph type to match whether direction, capacity or cost matters.

In a delivery network, a one-way road needs a directed edge and travel time needs a weight. Treating it as an unweighted undirected graph can permit an impossible or misleading route.

A graph edge is not automatically a physical straight line. State what vertices and edges mean before interpreting an algorithm's result.

Graph checklist: a simple graph has no loops or parallel edges; a complete graph joins every pair of distinct vertices; a subgraph uses selected vertices and edges; a tree is connected and has no cycles. In a directed graph distinguish in-degree from out-degree. Connected means every vertex pair is linked by a path in an undirected graph; strongly connected means directed paths exist in both directions between every pair.

An adjacency matrix records which vertices are directly connected. For an unweighted graph, Aᵢⱼ=1 means an edge from i to j under the chosen row/column convention.

The entry (A^k)ᵢⱼ counts walks of length k from i to j. Matrix multiplication works because each intermediate vertex is summed over, linking consecutive steps.

If two different two-step routes connect A to C, the corresponding entry of A² is 2. In a weighted matrix, the same multiplication may represent a different quantity, so do not mix conventions.

A walk may revisit vertices; it is not automatically a simple path. Check whether the graph is directed and what the matrix entries mean.

A weighted adjacency table stores a cost, distance or time rather than just 0 or 1. A transition matrix converts each vertex's outgoing weights or links into probabilities whose relevant row or column sums to 1 under the declared convention; repeated multiplication models movement through a strongly connected graph, as in a simplified PageRank model. Do not interpret powers of a weighted cost matrix as counts of walks unless that convention is explicitly justified.

A graph algorithm is a rule for extracting a route, assignment or bound from network data. A heuristic such as nearest neighbour chooses a locally short next edge; it is fast but not generally optimal.

Record the starting vertex and tie rule, then compare the resulting tour with a lower or upper bound when evaluating quality. Deleted-vertex or spanning-tree reasoning can provide a bound without proving the exact optimum.

A nearest-neighbour tour can choose the closest next city and later be forced into one very long final edge. A different early choice may produce a shorter total route, so the first route is a candidate, not a proof.

Greedy does not mean optimal. State the algorithm, its assumptions and whether the conclusion is a route, a bound or a proven minimum.

Algorithm map: an Eulerian trail uses every edge once (exactly two odd vertices) and an Eulerian circuit has all vertices even; Hamiltonian paths or cycles visit every vertex once. Kruskal selects globally smallest non-cycling edges, while Prim grows a minimum spanning tree from a chosen vertex. For a Chinese postman route, pair up to four odd vertices using least-distance paths, duplicate the minimum-total pairing, then take an Eulerian circuit. For a complete weighted travelling-salesman graph, nearest neighbour gives an upper bound and deleted-vertex plus MST reasoning gives a lower bound; complete a least-distance table first when the practical graph is not complete. State and justify the selected algorithm.