IB Maths AA HL · Unit 3: Geometry and Trigonometry
IB Maths AA HL Reciprocal and Inverse Trig Functions Questions
Exam-style IB Maths AA HL reciprocal and inverse trig functions questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.
- 32 questions
- Paper 1: 31
- Paper 2: 1
- 7 easy
- 11 medium
- 6 hard
- 8 starter
- 3 worked examples
Practise Reciprocal and Inverse Trig Functions questions →
AA HL formula booklet
What's examined in AA HL reciprocal and inverse trig functions
The question bank covers these reciprocal and inverse trig functions question types (number of questions in brackets):
- Inverse Trig Properties & Equations (12)
- Trigonometric Identities & Equations (11)
- Reciprocal Trig Manipulations (9)
Key formulas
- Compound-angle (cosine)
- \(\cos(A\pm B) = \cos A\cos B \mp \sin A\sin B\)
In the same notation as the IB formula booklet. All AA HL formulas →
Reciprocal and Inverse Trig Functions worked examples
Worked example 1: Evaluating reciprocal trig functions exactly · easy
Find the exact value of $\sec\left(\frac{\pi}{3}\right) + \cot\left(\frac{\pi}{4}\right)$.
1. Rewrite the reciprocal trigonometric functions in terms of the primary trigonometric functions: $\frac{1}{\cos(\pi/3)} + \frac{1}{\tan(\pi/4)}$.
2. Evaluate the exact value of the primary functions using the unit circle: $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$ and $\tan\left(\frac{\pi}{4}\right) = 1$.
3. Substitute these values back into the fractions: $\frac{1}{1/2} + \frac{1}{1}$.
4. Simplify the fractions: $2 + 1$.
5. State the final exact sum: $\mathbf{3}$.
Examiner tip: Remember that $\sec \theta$ is the reciprocal of $\cos \theta$, and $\csc \theta$ is the reciprocal of $\sin \theta$; mixing up which reciprocal pairs with which primary function is a very common and costly error.
Worked example 2: Solving equations with Pythagorean identities · medium
Solve the equation $\sec^2 x + 2\tan x = 4$ for $0 \le x \le \pi$. Give your answers in radians to three significant figures.
1. Identify the relevant Pythagorean identity linking secant and tangent from the formula booklet: $1 + \tan^2 x = \sec^2 x$.
2. Substitute this identity into the equation to create a uniform trigonometric function: $(1 + \tan^2 x) + 2\tan x = 4$.
3. Rearrange the terms to form a quadratic equation in terms of $\tan x$: $\tan^2 x + 2\tan x - 3 = 0$.
4. Factorise the quadratic expression: $(\tan x + 3)(\tan x - 1) = 0$, giving $\tan x = 1$ or $\tan x = -3$.
5. Solve the first case: $\tan x = 1 \implies x = \frac{\pi}{4} \approx 0.785$ radians.
6. Solve the second case using your GDC: $\arctan(-3) = -1.249...$ Since this is outside the domain $[0, \pi]$, add the period of tangent ($\pi$) to find the valid second-quadrant solution: $x = -1.249... + \pi = 1.892...$ Final answers: $\mathbf{x = 0.785, x = 1.89}$.
Examiner tip: Always verify your calculator output against the required domain; the GDC's principal value for $\arctan(-3)$ is negative, so you must manually add $\pi$ to shift it into the valid $0 \le x \le \pi$ interval.
Worked example 3: Evaluating inverse trigonometric compositions · hard
Find the exact value of $\sin\left(2\arccos\left(\frac{2}{3}\right)\right)$.
1. Assign a variable to the inner inverse trigonometric function to simplify the expression: Let $\theta = \arccos\left(\frac{2}{3}\right)$.
2. Rewrite this relationship using primary trigonometry: $\cos \theta = \frac{2}{3}$. Because the range of arccosine is $[0, \pi]$ and the value is positive, $\theta$ must be an acute angle in Quadrant 1.
3. Use the Pythagorean identity to find $\sin \theta$: $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{2}{3}\right)^2} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$.
4. Substitute $\theta$ back into the original expression, which now reads $\sin(2\theta)$.
5. Apply the double angle identity for sine: $\sin(2\theta) = 2\sin \theta \cos \theta$.
6. Calculate the final exact value by substituting the known fractions: $2\left(\frac{\sqrt{5}}{3}\right)\left(\frac{2}{3}\right) = \mathbf{\frac{4\sqrt{5}}{9}}$.
Examiner tip: Drawing a quick right-angled reference triangle (with adjacent side 2 and hypotenuse 3) is often the safest and most intuitive way to find the exact sine or tangent of an arccosine value without relying solely on memorized algebraic identities.
FAQ
How many IB Maths AA HL reciprocal and inverse trig functions questions are there?
There are 32 exam-style reciprocal and inverse trig functions questions in the AA HL question bank (Paper 1: 31 · Paper 2: 1), graded 7 easy, 11 medium, 6 hard, 8 starter. Every question has a full IB-style mark scheme (M, A and R marks).
Is reciprocal and inverse trig functions on Paper 1 or Paper 2?
Both. In the bank, Paper 1: 31 · Paper 2: 1. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.
Where can I get the mark schemes?
Open the AA HL Unit 3 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.
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