Geometry and Trigonometry — IB Math AA HL AA HL Study Guide

For: IB Math AA HL candidates sitting IB Math: Analysis & Approaches HL.

Covers: trigonometric identities and equations, unit circle exact values, sine, cosine and area rules for triangles, inverse trig function restricted domains, HL vector operations (dot and cross product), and 3D line and plane equations.

You should already know: IGCSE / pre-DP math, comfort with proof and algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Math AA HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Geometry and Trigonometry?

This topic unites spatial reasoning (geometry) and circular function analysis (trigonometry) to solve problems ranging from basic triangle side/angle calculations to 3D coordinate system navigation, a core HL syllabus component weighted at 15-20% of your final assessment. It combines algebraic manipulation of trigonometric expressions with vector arithmetic, and appears across all papers, including paper 3 investigative tasks. Common synonyms include circular functions and 3D coordinate geometry for its more advanced subcomponents.

2. Trig identities and equations

Trigonometric identities are true for all valid input values of the angle variable, and are used to simplify complex expressions, prove statements, and solve equations. The core identities you must memorize for HL are:

• Pythagorean identities: $${\sin }^2\theta +{\cos }^2\theta =11+{\tan }^2\theta ={\sec }^2\theta 1+{\cot }^2\theta ={\csc }^2\theta$$
• Compound angle identities: $$\sin (A\pm B)=\sin A\cos B\pm \cos A\sin B\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B$$ $$\tan (A\pm B)=\frac{\tan A\pm \tan B}{1\mp \tan A\tan B}$$
• Double angle identities (derived directly from compound angle identities where $A=B$): $$\sin 2\theta =2\sin \theta \cos \theta \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta =2{\cos }^2\theta -1=1-2{\sin }^2\theta$$

When solving trig equations, first rearrange using identities to isolate a single trigonometric function, then find all solutions in the given domain, accounting for periodicity ($\sin \theta ,\cos \theta$ have period $2\pi$; $\tan \theta$ has period $\pi$).

Worked Example: Solve $2{\cos }^{2}x=3\sin x$ for $0\le x\le 2\pi$.

1. Substitute ${\cos }^{2}x=1-{\sin }^{2}x$ to rewrite the equation in terms of $\sin x$: $$2(1-{\sin }^{2}x)=3\sin x⟹2{\sin }^{2}x+3\sin x-2=0$$
2. Let $y=\sin x$, solve the quadratic: $2y^2+3y-2=0$ gives $y=0.5$ or $y=-2$. Discard $y=-2$ as $\sin x$ cannot be less than -1.
3. Find solutions for $\sin x=0.5$ in $0\le x\le 2\pi$: $x=\frac{\pi }{6},\frac{5\pi }{6}$.

Exam tip: Examiners require explicit demonstration of identity substitution; skipping this step will lose method marks even if your final solutions are correct.

3. The unit circle and exact values

The unit circle is a circle of radius 1 centered at the origin, where for any angle $\theta$ measured counterclockwise from the positive x-axis, the coordinates of the intersection point are $(\cos \theta ,\sin \theta )$. This gives you a visual tool to calculate exact trig values for common angles and their multiples across all four quadrants, using the ASTC rule:

• All functions positive in Q1 ($0<\theta <\frac{\pi }{2}$)
• Sine positive in Q2 ($\frac{\pi }{2}<\theta <\pi$)
• Tangent positive in Q3 ($\pi <\theta <\frac{3\pi }{2}$)
• Cosine positive in Q4 ($\frac{3\pi }{2}<\theta <2\pi$)

Worked Example: Find the exact value of $\cos \left(\frac{5\pi }{3}\right)$.

1. $\frac{5\pi }{3}$ lies in Q4, where cosine is positive.
2. The reference angle (distance to the nearest x-axis) is $2\pi -\frac{5\pi }{3}=\frac{\pi }{3}$.
3. $\cos \left(\frac{\pi }{3}\right)=0.5$, so $\cos \left(\frac{5\pi }{3}\right)=0.5$.

Exam tip: Always confirm the quadrant before writing your exact value; this is the most common 1-mark loss on short-answer trig questions.

4. Sine, cosine and area rules

For non-right-angled triangles, label sides $a,b,c$ opposite angles $A,B,C$ respectively to apply the following rules:

• Sine Rule: Use when you have 2 angles and 1 side, or 2 sides and a non-included angle: $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$ where $R$ is the circumradius of the triangle. Note the ambiguous case: if you have 2 sides and a non-included acute angle, there may be 2 valid triangles.
• Cosine Rule: Use when you have 2 sides and an included angle, or 3 sides: $$a^2=b^2+c^2-2bc\cos A⟹\cos A=\frac{b^2+c^2-a^2}{2bc}$$
• Area Rule: Calculate area without needing the height of the triangle: $$Area=\frac{1}{2}ab\sin C$$

Worked Example: Triangle $ABC$ has $AB=5$cm, $BC=7$cm, $\angle B=60^{\circ }$. Find the area and length of $AC$.

1. Area: $\frac{1}{2}\times 5\times 7\times \sin 60^{\circ }=\frac{35\sqrt{3}}{4}\approx 15.2$ cm²
2. AC length (cosine rule): $A{C}^2=5^2+7^2-2(5)(7)\cos 60^{\circ }=74-35=39⟹AC=\sqrt{39}\approx 6.24$ cm.

5. Inverse trig functions — restricted domains

Standard trigonometric functions are not one-to-one over their full domains, so we restrict their domains to create invertible inverse functions that return unique principal values:

Worked Example: Find the exact value of $\arcsin \left(\sin \left(\frac{7\pi }{6}\right)\right)$.

1. First calculate $\sin \left(\frac{7\pi }{6}\right)=-\frac{1}{2}$.
2. $\arcsin \left(-\frac{1}{2}\right)$ must lie in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, so the value is $-\frac{\pi }{6}$, not $\frac{7\pi }{6}$ (which falls outside the principal range).

Exam tip: Examiners frequently test this "undoing" property; never cancel inverse and standard trig functions without verifying the original angle falls in the principal range.

6. Vectors — magnitude, direction, dot and cross products (HL)

Vectors have magnitude and direction, written as $\vec{a}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$ for 3D vectors. Key operations:

• Magnitude: $|\vec{a}|=\sqrt{x^2+y^2+z^2}$; unit vector in direction $\vec{a}$: $\hat{a}=\frac{\vec{a}}{|\vec{a}|}$
• Dot Product (scalar product): Measures the alignment of two vectors, zero if vectors are perpendicular: $$\vec{a}\cdot \vec{b}=x_1{x}_2+y_1{y}_2+z_1{z}_2=|\vec{a}||\vec{b}|\cos \theta$$ where $\theta$ is the angle between the vectors.
• Cross Product (vector product, HL only): Returns a vector perpendicular to both input vectors, zero if vectors are parallel: $$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|=|\vec{a}||\vec{b}|\sin \theta \hat{n}$$ The magnitude of $\vec{a}\times \vec{b}$ equals the area of the parallelogram formed by the two vectors.

Worked Example: For $\vec{a}=2\mathbf{i}+\mathbf{j}-3\mathbf{k}$ and $\vec{b}=\mathbf{i}-2\mathbf{j}+\mathbf{k}$, find the angle between them.

1. Dot product: $\vec{a}\cdot \vec{b}=2(1)+1(-2)+(-3)(1)=-3$
2. Magnitudes: $|\vec{a}|=\sqrt{4+1+9}=\sqrt{14}$, $|\vec{b}|=\sqrt{1+4+1}=\sqrt{6}$
3. $\cos \theta =\frac{-3}{\sqrt{14}\sqrt{6}}\approx -0.327$, so $\theta \approx 109^{\circ }$.

7. Equation of a line and plane in 3D (HL)

Line in 3D space

Three equivalent forms, where $\vec{a}$ is the position vector of a point on the line, and $\vec{d}$ is the direction vector:

1. Vector form: $\vec{r}=\vec{a}+t\vec{d}$, $t\in \mathbb{R}$
2. Parametric form: $x=a_x+t{d}_x,y=a_y+t{d}_y,z=a_z+t{d}_z$
3. Cartesian form: $\frac{x-a_x}{d_x}=\frac{y-a_y}{d_y}=\frac{z-a_z}{d_z}$

Plane in 3D space

Three equivalent forms, where $\vec{n}$ is the normal vector perpendicular to the plane:

1. Vector form: $\vec{r}\cdot \vec{n}=\vec{a}\cdot \vec{n}$
2. Cartesian form: $ax+by+cz=d$, where $\vec{n}=(a,b,c)$
3. Parametric form: $\vec{r}=\vec{a}+s\vec{u}+t\vec{v}$, where $\vec{u},\vec{v}$ are non-parallel vectors lying on the plane, $s,t\in \mathbb{R}$

Worked Example: Find the Cartesian equation of the plane passing through points $A(1,0,2)$, $B(2,1,1)$, $C(-1,2,1)$.

1. Find two vectors on the plane: $\overrightarrow{AB}=(1,1,-1)$, $\overrightarrow{AC}=(-2,2,-1)$
2. Calculate normal vector as cross product: $\overrightarrow{AB}\times \overrightarrow{AC}=(1,3,4)$
3. $d=\vec{n}\cdot \vec{a}=1(1)+3(0)+4(2)=9$, so Cartesian equation: $x+3y+4z=9$.

8. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting the ambiguous case of the sine rule, writing only one solution when two exist. Why students do it: Default assumption that all triangles are acute. Correct move: When working with 2 sides and a non-included acute angle, check if $180^{\circ }-\text{calculated angle}$ plus the given angle is less than $180^{\circ }$; if yes, include both solutions.
• Wrong move: Cancelling $\sin \theta$ from both sides of an equation, losing solutions where $\sin \theta =0$. Why students do it: Treating trig equations like linear equations, dividing without checking if the divisor can be zero. Correct move: Rearrange to factor out $\sin \theta$ instead of dividing, e.g., $\sin \theta \tan \theta =\sin \theta ⟹\sin \theta (\tan \theta -1)=0$.
• Wrong move: Mixing up dot and cross product properties, stating the cross product of perpendicular vectors is zero. Why students do it: Confusing scalar and vector product rules. Correct move: Remember dot product = 0 for perpendicular vectors, cross product = 0 for parallel vectors.
• Wrong move: Using the wrong principal range for inverse trig functions, e.g., writing $\frac{5\pi }{6}$ as the value of $\arcsin (0.5)$. Why students do it: Failing to memorize range differences between arcsin, arccos, arctan. Correct move: Write down the range for the relevant inverse function every time you solve an inverse trig question until you have it memorized.
• Wrong move: Marking a correct line equation as wrong because the direction vector is a scalar multiple of your calculated value. Why students do it: Fixating on one "correct" form of the line equation. Correct move: Verify equivalence by checking if the direction vector is a scalar multiple and the point given lies on the line.

9. Practice Questions (IB Math AA HL Style)

Question 1

Solve $\cos 2x+5\sin x=2$ for $0\le x\le 2\pi$, giving exact solutions.

Solution

1. Substitute $\cos 2x=1-2{\sin }^{2}x$: $$1-2{\sin }^{2}x+5\sin x=2⟹2{\sin }^{2}x-5\sin x+1=0$$
2. Solve quadratic for $\sin x$: $\sin x=\frac{5\pm \sqrt{25-8}}{4}=\frac{5\pm \sqrt{17}}{4}$
3. Discard $\frac{5+\sqrt{17}}{4}\approx 2.28$ (outside $\sin x$ range $[-1,1]$). Only valid solution: $\sin x=\frac{5-\sqrt{17}}{4}$
4. Exact solutions in domain: $x=\arcsin \left(\frac{5-\sqrt{17}}{4}\right),x=\pi -\arcsin \left(\frac{5-\sqrt{17}}{4}\right)$

Question 2

Triangle $PQR$ has $PQ=8$cm, $PR=12$cm, area = $24\sqrt{3}$ cm². Find the two possible measures of $\angle P$ and corresponding lengths of $QR$.

Solution

1. Use area rule: $24\sqrt{3}=\frac{1}{2}(8)(12)\sin P⟹\sin P=\frac{\sqrt{3}}{2}$
2. Two valid angles: $\angle P=60^{\circ }$ or $\angle P=120^{\circ }$
3. For $\angle P=60^{\circ }$: $Q{R}^2=8^2+12^2-2(8)(12)\cos 60^{\circ }=112⟹QR=4\sqrt{7}$ cm
4. For $\angle P=120^{\circ }$: $Q{R}^2=8^2+12^2-2(8)(12)\cos 120^{\circ }=304⟹QR=4\sqrt{19}$ cm

Question 3

Find the acute angle between line $L:\vec{r}=\left(\begin{array}{c}1\\ 2\\ 0\end{array}\right)+t\left(\begin{array}{c}2\\ 1\\ -2\end{array}\right)$ and plane $\Pi :3x-y+2z=5$.

Solution

1. The angle between a line and plane is $90^{\circ }$ minus the angle between the line's direction vector $\vec{d}=(2,1,-2)$ and the plane's normal vector $\vec{n}=(3,-1,2)$
2. Calculate dot product: $\vec{d}\cdot \vec{n}=2(3)+1(-1)+(-2)(2)=1$
3. Magnitudes: $|\vec{d}|=3$, $|\vec{n}|=\sqrt{14}$
4. Angle between $\vec{d}$ and $\vec{n}$: $\cos \varphi =\frac{1}{3\sqrt{14}}⟹\varphi \approx 84.9^{\circ }$
5. Acute angle between line and plane: $90^{\circ }-84.9^{\circ }=5.1^{\circ }$

10. Quick Reference Cheatsheet

11. What's Next

This topic forms the foundation for several later AA HL syllabus components, including calculus of trigonometric functions (differentiation and integration of sin, cos, tan, and inverse trig functions), complex numbers in polar and exponential form, and vector calculus applications in Paper 3 investigative tasks. A strong grasp of 3D line and plane equations is also required for problems involving intersections of geometric objects, which appear frequently in high-mark Paper 2 and Paper 3 questions.

If you struggle with any of the concepts, identities, or problem-solving techniques covered in this guide, you can ask Ollie for step-by-step explanations, additional practice questions, or custom quizzes tailored to your weak spots. Head to Ollie, the OwlsPrep AI tutor, to get personalised support as you prepare for your IB Math AA HL exams.