Identities & Equations (AA HL)
A trigonometric identity is true for every value of the angle involved, while a trigonometric equation is only true for specific values you have to find. This topic builds the toolkit of identities - Pythagorean, compound-angle and double-angle - that let you rewrite one trig expression as another, and then uses that toolkit to solve equations exactly, without a calculator, over a stated interval.
What the syllabus says
This topic maps onto four points in the official IB Analysis & Approaches syllabus, two shared with SL and two exclusive to HL.
| Code | Syllabus content |
|---|---|
| SL3.6 | The Pythagorean identity \(\cos^2\theta+\sin^2\theta=1\). Double angle identities for sine and cosine. |
| SL3.8 | Solving trigonometric equations in a finite interval, both graphically and analytically, including equations leading to quadratic equations in \(\sin x\), \(\cos x\) or \(\tan x\). |
| AHL3.9 | Definition of the reciprocal trigonometric ratios \(\sec\theta\), \(\csc\theta\) and \(\cot\theta\). Pythagorean identities \(1+\tan^2\theta=\sec^2\theta\) and \(1+\cot^2\theta=\csc^2\theta\). The inverse functions \(\arcsin x\), \(\arccos x\), \(\arctan x\); their domains, ranges and graphs. |
| AHL3.10 | Compound angle identities. Derivation of the double angle identities from the compound angle identities. Double angle identity for \(\tan\theta\). |
SL3.6 and SL3.8 are also examinable at AA HL; AHL3.9 and AHL3.10 extend the toolkit with reciprocal ratios and compound angles.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is a Pythagorean identity?
A Pythagorean identity relates \(\sin\theta\), \(\cos\theta\) and (at HL) their reciprocals, coming from the fact that a point on the unit circle satisfies \(x^2+y^2=1\). It lets you find one ratio from another, up to a sign you must determine from the quadrant.
e.g. \(\sin^2 30^\circ+\cos^2 30^\circ = \left(\tfrac12\right)^2+\left(\tfrac{\sqrt3}{2}\right)^2 = \tfrac14+\tfrac34 = 1\).
What is a double angle identity?
A double angle identity rewrites a trig function of \(2\theta\) in terms of functions of \(\theta\) alone, e.g. \(\cos2\theta=1-2\sin^2\theta\). It's the special case of a compound angle identity where both angles are equal.
e.g. \(\cos60^\circ = 1-2\sin^2 30^\circ = 1-2\left(\tfrac14\right) = \tfrac12\).
What is a compound angle identity?
A compound angle identity expands a trig function of a sum or difference of two angles, e.g. \(\sin(A+B)=\sin A\cos B+\cos A\sin B\). It's how you find exact values for angles like \(75^\circ\) that aren't on the standard unit circle.
e.g. \(\sin75^\circ=\sin(45^\circ+30^\circ)=\tfrac{\sqrt2}{2}\cdot\tfrac{\sqrt3}{2}+\tfrac{\sqrt2}{2}\cdot\tfrac12=\dfrac{\sqrt6+\sqrt2}{4}\approx0.966\).
What are the reciprocal trigonometric ratios?
The reciprocal ratios flip sine, cosine and tangent upside down: \(\sec\theta=\tfrac{1}{\cos\theta}\), \(\csc\theta=\tfrac{1}{\sin\theta}\), \(\cot\theta=\tfrac{1}{\tan\theta}\). They come with their own Pythagorean identities, built by dividing the main one by \(\sin^2\theta\) or \(\cos^2\theta\).
e.g. \(\sec60^\circ = \dfrac{1}{\cos60^\circ} = \dfrac{1}{0.5} = 2\).
What does it mean to solve a trig equation in a finite interval?
Trig functions repeat, so an equation like \(\sin x = k\) has infinitely many solutions overall. "Solve in a finite interval" means listing only the solutions that fall inside a given range, such as \(0\le x\le2\pi\).
e.g. \(2\sin x=1\) for \(0\le x\le2\pi\) gives \(x=\tfrac{\pi}{6}\) and \(x=\pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}\).
Key formulas
Seven formulas cover almost every question on this topic. The two tables below summarise all of them at a glance - the explanations underneath go into more depth on each one.
Formula reference
Every identity below is printed in the official formula booklet, under the geometry and trigonometry section.
| Formula | Used for | Booklet? |
|---|---|---|
| \(\cos^2\theta+\sin^2\theta=1\) | Pythagorean identity | ✓ Yes |
| \(1+\tan^2\theta=\sec^2\theta\) | Pythagorean identity (tan/sec) | ✓ Yes |
| \(\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\) | Compound angle (sine) | ✓ Yes |
| \(\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\) | Compound angle (cosine) | ✓ Yes |
| \(\sin2\theta=2\sin\theta\cos\theta\) | Double angle (sine) | ✓ Yes |
| \(\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta\) | Double angle (cosine, three forms) | ✓ Yes |
| \(\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}\) | Double angle (tan) | ✓ Yes |
Choosing the right form of \(\cos2\theta\)
All three forms of the cosine double angle identity are equivalent - the trick is picking the one that avoids an extra step, based on what you already know.
| Form | Use when |
|---|---|
| \(\cos2\theta=\cos^2\theta-\sin^2\theta\) | You already know both \(\sin\theta\) and \(\cos\theta\). |
| \(\cos2\theta=2\cos^2\theta-1\) | You only know \(\cos\theta\). |
| \(\cos2\theta=1-2\sin^2\theta\) | You only know \(\sin\theta\). |
Pythagorean identities
Each identity comes from dividing \(\cos^2\theta+\sin^2\theta=1\) through by \(\cos^2\theta\) or \(\sin^2\theta\).
Main identity
\[\cos^2\theta+\sin^2\theta=1\]
Find \(\sin\theta\) from \(\cos\theta\), or vice versa - then fix the sign using the quadrant.
✓ In the formula bookletTan/sec identity
\[1+\tan^2\theta=\sec^2\theta\]
Divide the main identity by \(\cos^2\theta\); useful whenever \(\sec\theta\) or \(\tan\theta\) appears.
✓ In the formula bookletCot/cosec identity
\[1+\cot^2\theta=\csc^2\theta\]
Divide the main identity by \(\sin^2\theta\); useful whenever \(\csc\theta\) or \(\cot\theta\) appears.
✓ In the formula bookletCompound and double angle identities
The double angle identities are just the compound angle identities with \(A=B=\theta\).
Compound angle (sine)
\[\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\]
Use for exact values of angles built from two known angles, like \(75^\circ=45^\circ+30^\circ\).
✓ In the formula bookletCompound angle (cosine)
\[\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\]
Note the sign flips: \(+\) in the identity pairs with \(-\) inside the cosines, and vice versa.
✓ In the formula bookletDouble angle (tan)
\[\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}\]
Comes from dividing \(\sin2\theta\) by \(\cos2\theta\) and simplifying; undefined when \(\tan\theta=\pm1\).
✓ In the formula bookletSolving trigonometric equations
Most equation questions reduce to a simpler equation in a single trig ratio, then use the identities above to find every solution in range.
Reduce to one ratio
Use an identity to rewrite the equation so only one trig function of one angle appears, e.g. turn \(\cos2\theta\) into an expression in \(\sin\theta\) alone.
Solve like a quadratic
If the identity introduces a squared term, the equation is often quadratic in \(\sin\theta\) (or \(\cos\theta\)) - factorise or use the quadratic formula, letting \(u=\sin\theta\).
Check the interval
Find the general solution first, then list only the values that fall inside the given interval - it's easy to drop a valid solution or include one that's out of range.
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
Given \(\cos\theta=-\tfrac{5}{13}\) with \(\tfrac{\pi}{2}<\theta<\pi:\)
(a) Find the exact value of \(\sin\theta.\)
(b) Find the exact value of \(\tan\theta.\)
(c) Find the exact value of \(\sec\theta.\)
Worked solution
(a) \(\sin^2\theta=1-\tfrac{25}{169}=\tfrac{144}{169}.\) M1
As \(\theta\) is in quadrant II, \(\sin\theta=\tfrac{12}{13}.\) A1
(b) \(\tan\theta=\dfrac{12/13}{-5/13}=-\tfrac{12}{5}.\) A1
(c) \(\sec\theta=\dfrac1{\cos\theta}=-\dfrac{13}{5}.\) A1
\(\tan\theta=-3.\)
(a) Find the exact value of \(\tan2\theta.\)
(b) Hence find the exact value of \(\tan4\theta.\)
Worked solution
(a) \(\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}=\dfrac{2(-3)}{1-9}\) M1
\(=\dfrac{-6}{-8}=\tfrac34.\) A1
(b) \(\tan4\theta=\dfrac{2\tan2\theta}{1-\tan^22\theta}=\dfrac{2(3/4)}{1-9/16}=\dfrac{3/2}{7/16}\) M1
\(=\dfrac{24}{7}.\) A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Forgetting to check the quadrant sign. The Pythagorean identity only gives you \(\sin^2\theta\) or \(\cos^2\theta\) - you must use the given range of \(\theta\) to decide whether the square root is positive or negative.
- Losing track of which form of \(\cos2\theta\) to use. Mixing up the three equivalent forms wastes time converting between \(\sin\theta\) and \(\cos\theta\) when you already had the one you needed.
- Splitting \(\sin(A+B)\) into \(\sin A+\sin B\). Sine and cosine don't distribute over addition - the compound angle formula has four terms, not two.
- Dropping solutions when solving in an interval. After finding one solution, always check the other cases from the identity (e.g. both roots of a quadratic in \(\sin\theta\)) and make sure every valid angle in the interval is listed.
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
Faster and safer than algebra for messy equations - and the trick is getting all solutions, not just one.
- Graph \(f(x)\) first so you can see how many solutions exist and roughly where they are.
- Rearrange so everything is on one side: \(f(x) = 0\) - or graph both sides as separate functions and find intersections.
- MATH → Solver: enter the expression, type a starting guess close to one root, press ALPHA + ENTER. Move the guess to near a different root and repeat for each solution.TI-84
- Type nSolve(f(x)=0, x, guess) - include a guess or interval e.g. nSolve(f(x)=0, x, 2) or nSolve(f(x)=0, x, {1,5}) to target a specific root.Nspire
- Run-Matrix → SolveN(f(x), x) returns all real roots at once; or use the Equation app for a visual approach.Casio
- For equations like \(\sin x = 0.5\) over an interval, graph both sides and use the intersection tool rather than the equation solver - it's faster and less likely to miss roots.
- Always verify each solution by substituting back into the original equation.
Tip: The solver finds ONE root near your starting guess - change the guess to find others. The graph shows you how many to expect.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Identities & equations questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What's the difference between an identity and an equation?
An identity is true for every value of the variable, and is proved by rearranging one side until it matches the other. An equation is only true for specific values, which you solve for - usually within a stated interval.
When do I use the compound angle formulas instead of the double angle formulas?
The double angle formulas are just the compound angle formulas with \(A=B=\theta\). Use compound angle when you're combining two different angles, like \(\tan15^\circ=\tan(45^\circ-30^\circ)\). Use double angle when you're given information about a single angle \(\theta\) and need an expression in \(2\theta\).
How do I know which form of \(\cos2\theta\) to use?
Use \(\cos^2\theta-\sin^2\theta\) if you know both \(\sin\theta\) and \(\cos\theta\). Use \(2\cos^2\theta-1\) if you only know \(\cos\theta\). Use \(1-2\sin^2\theta\) if you only know \(\sin\theta\). Picking the right form avoids an extra step.
Are trigonometric equations examined without a calculator?
Yes. Paper 1 regularly asks you to solve trig equations using exact values and identities, without technology. Paper 2 allows you to confirm solutions graphically on your GDC. See the GDC guide for model-specific instructions.
Sub-topics
Identities & Equations broken down into its individual skills, each with its own focused page.
Related topics
More Geometry & Trigonometry topics from the same AA HL syllabus unit, in case you want to keep going.