How do I remember the signs in the compound-angle formulae?

For $\sin(A\pm B)$ the sign inside the expansion is the SAME as the sign between the angles ($\sin A\cos B\pm\cos A\sin B$). For $\cos(A\pm B)$ the sign FLIPS ($\cos A\cos B\mp\sin A\sin B$): 'cosine changes the sign'. They are printed in the formula booklet, but recalling them instantly saves time in Paper 1.

Where do the double-angle formulae come from?

Set $B=A$ in the compound-angle formulae. $\sin 2\theta=\sin(\theta+\theta)=\sin\theta\cos\theta+\cos\theta\sin\theta=2\sin\theta\cos\theta$. Similarly $\cos 2\theta=\cos\theta\cos\theta-\sin\theta\sin\theta=\cos^2\theta-\sin^2\theta$, and substituting the Pythagorean identity gives the other two cosine forms. Knowing the derivation means you never need to memorise the doubles separately.

Which of the three $\cos 2\theta$ forms should I use?

Match the form to the rest of the equation so everything ends up in ONE function. If the other terms are in $\sin$, use $\cos 2\theta=1-2\sin^2\theta$. If they are in $\cos$, use $\cos 2\theta=2\cos^2\theta-1$. When proving an identity, the raw form $\cos^2\theta-\sin^2\theta$ is often the most flexible.

Why are the ranges of the inverse trig functions restricted?

$\sin$, $\cos$ and $\tan$ are periodic, so infinitely many angles share the same value; a function must give exactly one output per input. We therefore restrict each trig function to an interval on which it is one-to-one — $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$ for sine, $[0,\pi]$ for cosine, $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ for tangent — and that interval becomes the range of principal values for the inverse.

Why is $\arctan$'s range OPEN while $\arcsin$ and $\arccos$ are closed?

$\tan$ heads to $\pm\infty$ as the angle approaches $\pm\tfrac{\pi}{2}$ but never actually reaches those angles, so $\pm\tfrac{\pi}{2}$ are horizontal asymptotes of $y=\arctan x$ and are excluded: the range is the OPEN interval $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$. Sine and cosine attain their extreme values at the ends of their restricted domains, so those endpoints are included and the ranges are closed.

How do I evaluate something like $\cos(\arctan 2)$?

Let $\theta=\arctan 2$, so $\tan\theta=2$ with $\theta$ in $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$. Draw a right triangle with opposite $2$ and adjacent $1$; the hypotenuse is $\sqrt{2^2+1^2}=\sqrt5$. Then $\cos\theta=\dfrac{1}{\sqrt5}=\dfrac{\sqrt5}{5}$. Because $\arctan$ returns a Q1/Q4 angle, choose the sign of the cosine to match that quadrant (here positive).

Compound and double angle identities;…

Revise Compound and double angle identities; inverse trig for IB Mathematics: Analysis and Approaches (3.6)

Looking for free IB Mathematics: Analysis and Approaches notes on Compound and double angle identities; inverse trig? This Higher Level lesson covers official syllabus point 3.6 on Maths AA HL. Use the visual steps, key terms, and quick checks below — then practise with IB criterion marking on Compound and double angle identities; inverse trig. Built for students searching "IB Maths AA 3.6", "Compound and double angle identities; inverse trig IB revision", or a free alternative to paid IB note sites.

Compound and double angle identities; inverse trig

You already know the exact ratios for $30^\circ$ , $45^\circ$ and $60^\circ$ . But what about $75^\circ$ , or $\sin 2\theta$ when only $\sin\theta$ is known? It is tempting to write $\cos(30^\circ+45^\circ)=\cos 30^\circ+\cos 45^\circ$ , but that is simply false — trig functions do not distribute over addition. This lesson supplies the identities that govern how angles combine: the compound-angle formulae, the double-angle formulae they generate, and the inverse functions that recover an angle from a ratio. As in every calculation topic, the worked examples are laid out exactly the way our engine marks your own practice: the method is on the page, mark by mark.

BY THE END, YOU CAN…

1Recall and apply the compound-angle identities for $\sin(A\pm B)$ , $\cos(A\pm B)$ and $\tan(A\pm B)$ .
2Derive the double-angle identities $\sin 2\theta=2\sin\theta\cos\theta$ , $\cos 2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1$ and $\tan 2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$ from the compound-angle formulae.
3Use these identities to evaluate exact values, prove identities and solve trigonometric equations over a given interval.
4Define $\arcsin$ , $\arccos$ and $\arctan$ , stating their restricted domains, principal-value ranges and the shape of their graphs.
5Present working so that method marks and follow-through (FT) protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Plus-One and Times-Two Rules of Trigonometry

The compound-angle identities tell you the sine, cosine or tangent of a SUM or DIFFERENCE of two angles in terms of the angles separately. Set the two angles equal and they collapse into the double-angle identities. The inverse trig functions run the machinery backwards: given a ratio, they return the one principal angle that produced it.

THINK OF IT LIKE…

Think of an angle as a recipe and its sine as the finished dish. The compound-angle formula is the rule for combining two recipes into one; the double-angle formula is what happens when you cook the same recipe twice. The inverse functions are the food critic working in reverse — tasting the dish and naming the single 'principal' recipe behind it. Because many recipes can give the same taste, the critic is only allowed to name one, which is exactly why the ranges of $\arcsin$ , $\arccos$ and $\arctan$ are restricted.

1
Spot the structure: a sum/difference of angles ( $A\pm B$ ), a doubled angle ( $2\theta$ ), or an inverse function returning an angle.
2
Choose the matching identity from the formula booklet — for $\cos 2\theta$ , pick the form that leaves the equation in a single function.
3
Substitute known exact values carefully, keeping surds and fractions rather than decimals in Paper 1.
4
For inverse functions, give the ONE angle inside the principal-value range; for equations, collect EVERY solution in the stated interval.

Explore the concept

Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.

GEOGEBRASTEP 1 / 4 · Step 1

Loading simulation…

Fetching from GeoGebra…

Visualising Compound Angle Identities

Use “Visualising Compound Angle Identities” — shift curves or drag points to see equilibrium and slopes change.

If the embed does not load, use Open on GeoGebra.

GeoGebra · GeoGebra Terms of Service

Step 1

Spot the structure: a sum/difference of angles ( $A\pm B$ ), a doubled angle ( $2\theta$ ), or an inverse function returning an angle.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\\ \\ \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\\ \\ \tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}$

$\sin 2\theta=2\sin\theta\cos\theta\\ \\ \cos 2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1\\ \\ \tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$

Full topic notes

Formal explanation with the rigour you need for the exam.

The compound-angle (addition) identities

The compound-angle identities express the sine, cosine or tangent of a SUM or DIFFERENCE of two angles in terms of the ratios of the angles separately. They are provided in the formula booklet, so the marks come not from recall but from fluent, accurate application — especially for exact values in Paper 1.

$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\\ \\ \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\\ \\ \tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}$

Sign rule: for $\sin$ the inside sign MATCHES the sign between the angles; for $\cos$ it FLIPS ('cosine changes the sign').
Read the paired $\pm$ / $\mp$ carefully: in $\cos(A+B)$ the middle term is a MINUS, and in $\tan$ the denominator sign is opposite to the numerator.
These formulae unlock exact values for non-standard angles built from standard ones: $15^\circ=45^\circ-30^\circ$ , $75^\circ=45^\circ+30^\circ$ , $105^\circ=60^\circ+45^\circ$ .
The tangent form follows by dividing $\sin(A\pm B)$ by $\cos(A\pm B)$ and then dividing numerator and denominator by $\cos A\cos B$ .

The double-angle identities

Setting $B=A$ in the compound-angle formulae produces the double-angle identities. They let you collapse a doubled angle into single-angle terms, which is essential for solving equations and reappears throughout calculus. The cosine version has three equivalent forms, obtained by substituting the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$ .

$\sin 2\theta=2\sin\theta\cos\theta\\ \\ \cos 2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1\\ \\ \tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$

Derivation: $\sin 2\theta=\sin(\theta+\theta)=\sin\theta\cos\theta+\cos\theta\sin\theta=2\sin\theta\cos\theta$ ; likewise $\cos 2\theta=\cos^2\theta-\sin^2\theta$ .
The three cosine forms are equal: replace $\cos^2\theta$ with $1-\sin^2\theta$ to get $1-2\sin^2\theta$ ; replace $\sin^2\theta$ with $1-\cos^2\theta$ to get $2\cos^2\theta-1$ .
Choosing the RIGHT cosine form is the whole skill: pick the one matching the other terms so the equation reduces to a single function.
Common slip: $\sin 2\theta\neq 2\sin\theta$ and $\cos 2\theta\neq 2\cos\theta$ . The identities involve products and squares, not simple doublings.

The inverse trigonometric functions

The inverse functions $\arcsin x$ , $\arccos x$ and $\arctan x$ answer the question 'which angle has this ratio?'. Because $\sin$ , $\cos$ and $\tan$ are periodic and therefore many-to-one, they have no inverse over their whole domains. We restrict each to an interval on which it is one-to-one; that restricted domain becomes the RANGE of principal values for the inverse. These ranges are not printed in the formula booklet — you must know them.

$y=\arcsin x$ : domain $x\in[-1,1]$ , range $y\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ (Quadrants 4 and 1). The graph is increasing, passing through the origin.
$y=\arccos x$ : domain $x\in[-1,1]$ , range $y\in[0,\pi]$ (Quadrants 1 and 2). The graph is decreasing from $(-1,\pi)$ to $(1,0)$ .
$y=\arctan x$ : domain $x\in\mathbb{R}$ , range $y\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ — OPEN interval, with horizontal asymptotes $y=\pm\tfrac{\pi}{2}$ that the increasing graph approaches but never meets.
Each inverse graph is the reflection of the (restricted) original in the line $y=x$ ; reflecting swaps domain and range.

Common mistakes examiners penalise

Treating trig as distributive — $\sin(A+B)\neq\sin A+\sin B$ and $\cos(A+B)\neq\cos A+\cos B$ . Always expand with the compound-angle formula.
Getting the compound signs backwards — $\cos(A+B)$ uses a MINUS ( $\cos A\cos B-\sin A\sin B$ ) while $\cos(A-B)$ uses a plus. For sine it is the other way round. 'Cosine changes the sign' is the rescue.
Writing $\sin 2\theta=2\sin\theta$ or $\cos 2\theta=2\cos\theta$ — the double-angle identities are $\sin 2\theta=2\sin\theta\cos\theta$ and $\cos 2\theta=\cos^2\theta-\sin^2\theta$ , not simple doublings.
Choosing the wrong form of $\cos 2\theta$ — pick the form that leaves the equation in ONE function ( $1-2\sin^2\theta$ when the rest is in $\sin$ ; $2\cos^2\theta-1$ when it is in $\cos$ ). The wrong choice leaves a mixture you cannot factorise.
Dividing an equation by $\sin x$ or $\cos x$ — this assumes it is non-zero and silently LOSES the solutions where it is zero (e.g. dividing $\sin 2x=\sin x$ by $\sin x$ drops $\sin x=0$ ). Factorise instead.
Confusing inverse ranges — $\arcsin$ / $\arctan$ return Q4/Q1 angles; $\arccos$ returns Q1/Q2 (its value can be obtuse). Quoting the wrong-quadrant angle loses the accuracy mark.
Using closed brackets for $\arctan$ — its range is the OPEN interval $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ ; the endpoints are asymptotes and are never attained.

Model answer — marked the way our engine marks it

In Paper 1 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an accuracy mark (A) — and accuracy marks are DEPENDENT on the associated method mark. Follow-through (FT) means an early slip need not cost the marks that follow, provided your later work is correct on your own figures, and the engine accepts equivalent and exact forms ( $\tfrac{24}{25}$ , $0.96$ ). But this protection only exists if your method is on the page. Study how each mark below is earned by a specific line.

Where this leads

The compound- and double-angle identities are the workhorses of the rest of trigonometry and calculus. The double-angle results let you rewrite $\sin^2 x$ and $\cos^2 x$ before integrating, and reappear when you differentiate trig functions or simplify expressions ahead of finding areas and volumes of revolution. The inverse functions define the antiderivatives that produce $\arcsin$ , $\arccos$ and $\arctan$ , and underpin solving equations where you must recover the principal angle. Master the habit — pick the identity that reduces to a single function, keep exact forms, choose the sign from the quadrant, and show every line — and the harder trigonometry becomes variations on a method you already own.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

Without a calculator, find the exact value of $\cos 75^\circ$ . [4]

1
Write $75^\circ$ as a sum of standard angles: $75^\circ=45^\circ+30^\circ$ , and use $\cos(A+B)=\cos A\cos B-\sin A\sin B$ .

EXAMPLE 2

Prove the identity $\dfrac{\sin 2\theta}{1+\cos 2\theta}\equiv\tan\theta$ . [4]

1
Start with the left-hand side and choose double-angle forms that expose a common factor. For $\cos 2\theta$ use $2\cos^2\theta-1$ , so the denominator simplifies.

EXAMPLE 3

Find the exact value of $\sin\left(\arccos\left(-\tfrac{1}{3}\right)\right)$ . [4]

1
Let $\theta=\arccos\!\left(-\tfrac13\right)$ , so $\cos\theta=-\tfrac13$ with $\theta\in[0,\pi]$ by the range of $\arccos$ . (M1) for setting up the angle and stating the range.

EXAMPLE 4

Given that $\sin\theta = \dfrac{3}{5}$ and $\theta$ is acute, find the exact values of $\sin 2\theta$ and $\cos 2\theta$ . [5]

1
Model answer — full working.

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEACompound and double angle identities; inverse trig

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 12

Compound angle: $\sin(A\pm B)$

TAP OR SPACE TO FLIP

ANSWER

$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ . The sign inside is the SAME as the sign between the angles.

TAP OR SPACE TO FLIP BACK

1 / 12 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
Sign rule: for $\sin$ the inside sign MATCHES the sign between the angles; for $\cos$ it FLIPS ('cosine changes the sign').
✓
Read the paired $\pm$ / $\mp$ carefully: in $\cos(A+B)$ the middle term is a MINUS, and in $\tan$ the denominator sign is opposite to the numerator.
✓
These formulae unlock exact values for non-standard angles built from standard ones: $15^\circ=45^\circ-30^\circ$ , $75^\circ=45^\circ+30^\circ$ , $105^\circ=60^\circ+45^\circ$ .
✓
The tangent form follows by dividing $\sin(A\pm B)$ by $\cos(A\pm B)$ and then dividing numerator and denominator by $\cos A\cos B$ .

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Get a Paper 1 solution marked: find exact values with the double-angle identities

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Khan Academy: Inverse Trigonometric Functions↗

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Get a Paper 1 solution marked: find exact values with the double-angle identities on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS3.5 Trigonometric identities and equations NEXT →3.7 Vectors: properties and operations

Compound and double angle identities; inverse trig

BY THE END, YOU CAN…

1Recall and apply the compound-angle identities for $\sin(A\pm B)$ , $\cos(A\pm B)$ and $\tan(A\pm B)$ .
2Derive the double-angle identities $\sin 2\theta=2\sin\theta\cos\theta$ , $\cos 2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1$ and $\tan 2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$ from the compound-angle formulae.
3Use these identities to evaluate exact values, prove identities and solve trigonometric equations over a given interval.
4Define $\arcsin$ , $\arccos$ and $\arctan$ , stating their restricted domains, principal-value ranges and the shape of their graphs.
5Present working so that method marks and follow-through (FT) protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Plus-One and Times-Two Rules of Trigonometry

THINK OF IT LIKE…

1
Spot the structure: a sum/difference of angles ( $A\pm B$ ), a doubled angle ( $2\theta$ ), or an inverse function returning an angle.
2
Choose the matching identity from the formula booklet — for $\cos 2\theta$ , pick the form that leaves the equation in a single function.
3
Substitute known exact values carefully, keeping surds and fractions rather than decimals in Paper 1.
4
For inverse functions, give the ONE angle inside the principal-value range; for equations, collect EVERY solution in the stated interval.

Explore the concept

Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.

GEOGEBRASTEP 1 / 4 · Step 1

Loading simulation…

Fetching from GeoGebra…

Visualising Compound Angle Identities

Use “Visualising Compound Angle Identities” — shift curves or drag points to see equilibrium and slopes change.

If the embed does not load, use Open on GeoGebra.

GeoGebra · GeoGebra Terms of Service

Step 1

Spot the structure: a sum/difference of angles ( $A\pm B$ ), a doubled angle ( $2\theta$ ), or an inverse function returning an angle.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\\ \\ \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\\ \\ \tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}$

$\sin 2\theta=2\sin\theta\cos\theta\\ \\ \cos 2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1\\ \\ \tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$

Full topic notes

Formal explanation with the rigour you need for the exam.

The compound-angle (addition) identities

$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\\ \\ \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\\ \\ \tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}$

Sign rule: for $\sin$ the inside sign MATCHES the sign between the angles; for $\cos$ it FLIPS ('cosine changes the sign').
Read the paired $\pm$ / $\mp$ carefully: in $\cos(A+B)$ the middle term is a MINUS, and in $\tan$ the denominator sign is opposite to the numerator.
These formulae unlock exact values for non-standard angles built from standard ones: $15^\circ=45^\circ-30^\circ$ , $75^\circ=45^\circ+30^\circ$ , $105^\circ=60^\circ+45^\circ$ .
The tangent form follows by dividing $\sin(A\pm B)$ by $\cos(A\pm B)$ and then dividing numerator and denominator by $\cos A\cos B$ .

The double-angle identities

$\sin 2\theta=2\sin\theta\cos\theta\\ \\ \cos 2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1\\ \\ \tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$

Derivation: $\sin 2\theta=\sin(\theta+\theta)=\sin\theta\cos\theta+\cos\theta\sin\theta=2\sin\theta\cos\theta$ ; likewise $\cos 2\theta=\cos^2\theta-\sin^2\theta$ .
The three cosine forms are equal: replace $\cos^2\theta$ with $1-\sin^2\theta$ to get $1-2\sin^2\theta$ ; replace $\sin^2\theta$ with $1-\cos^2\theta$ to get $2\cos^2\theta-1$ .
Choosing the RIGHT cosine form is the whole skill: pick the one matching the other terms so the equation reduces to a single function.
Common slip: $\sin 2\theta\neq 2\sin\theta$ and $\cos 2\theta\neq 2\cos\theta$ . The identities involve products and squares, not simple doublings.

The inverse trigonometric functions

$y=\arcsin x$ : domain $x\in[-1,1]$ , range $y\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ (Quadrants 4 and 1). The graph is increasing, passing through the origin.
$y=\arccos x$ : domain $x\in[-1,1]$ , range $y\in[0,\pi]$ (Quadrants 1 and 2). The graph is decreasing from $(-1,\pi)$ to $(1,0)$ .
$y=\arctan x$ : domain $x\in\mathbb{R}$ , range $y\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ — OPEN interval, with horizontal asymptotes $y=\pm\tfrac{\pi}{2}$ that the increasing graph approaches but never meets.
Each inverse graph is the reflection of the (restricted) original in the line $y=x$ ; reflecting swaps domain and range.

Common mistakes examiners penalise

Treating trig as distributive — $\sin(A+B)\neq\sin A+\sin B$ and $\cos(A+B)\neq\cos A+\cos B$ . Always expand with the compound-angle formula.
Getting the compound signs backwards — $\cos(A+B)$ uses a MINUS ( $\cos A\cos B-\sin A\sin B$ ) while $\cos(A-B)$ uses a plus. For sine it is the other way round. 'Cosine changes the sign' is the rescue.
Writing $\sin 2\theta=2\sin\theta$ or $\cos 2\theta=2\cos\theta$ — the double-angle identities are $\sin 2\theta=2\sin\theta\cos\theta$ and $\cos 2\theta=\cos^2\theta-\sin^2\theta$ , not simple doublings.
Choosing the wrong form of $\cos 2\theta$ — pick the form that leaves the equation in ONE function ( $1-2\sin^2\theta$ when the rest is in $\sin$ ; $2\cos^2\theta-1$ when it is in $\cos$ ). The wrong choice leaves a mixture you cannot factorise.
Dividing an equation by $\sin x$ or $\cos x$ — this assumes it is non-zero and silently LOSES the solutions where it is zero (e.g. dividing $\sin 2x=\sin x$ by $\sin x$ drops $\sin x=0$ ). Factorise instead.
Confusing inverse ranges — $\arcsin$ / $\arctan$ return Q4/Q1 angles; $\arccos$ returns Q1/Q2 (its value can be obtuse). Quoting the wrong-quadrant angle loses the accuracy mark.
Using closed brackets for $\arctan$ — its range is the OPEN interval $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ ; the endpoints are asymptotes and are never attained.

Model answer — marked the way our engine marks it

Where this leads

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

Without a calculator, find the exact value of $\cos 75^\circ$ . [4]

1
Write $75^\circ$ as a sum of standard angles: $75^\circ=45^\circ+30^\circ$ , and use $\cos(A+B)=\cos A\cos B-\sin A\sin B$ .

EXAMPLE 2

Prove the identity $\dfrac{\sin 2\theta}{1+\cos 2\theta}\equiv\tan\theta$ . [4]

1
Start with the left-hand side and choose double-angle forms that expose a common factor. For $\cos 2\theta$ use $2\cos^2\theta-1$ , so the denominator simplifies.

EXAMPLE 3

Find the exact value of $\sin\left(\arccos\left(-\tfrac{1}{3}\right)\right)$ . [4]

1
Let $\theta=\arccos\!\left(-\tfrac13\right)$ , so $\cos\theta=-\tfrac13$ with $\theta\in[0,\pi]$ by the range of $\arccos$ . (M1) for setting up the angle and stating the range.

EXAMPLE 4

Given that $\sin\theta = \dfrac{3}{5}$ and $\theta$ is acute, find the exact values of $\sin 2\theta$ and $\cos 2\theta$ . [5]

1
Model answer — full working.

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEACompound and double angle identities; inverse trig

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 12

Compound angle: $\sin(A\pm B)$

TAP OR SPACE TO FLIP

ANSWER

$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ . The sign inside is the SAME as the sign between the angles.

TAP OR SPACE TO FLIP BACK

1 / 12 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
Sign rule: for $\sin$ the inside sign MATCHES the sign between the angles; for $\cos$ it FLIPS ('cosine changes the sign').
✓
Read the paired $\pm$ / $\mp$ carefully: in $\cos(A+B)$ the middle term is a MINUS, and in $\tan$ the denominator sign is opposite to the numerator.
✓
These formulae unlock exact values for non-standard angles built from standard ones: $15^\circ=45^\circ-30^\circ$ , $75^\circ=45^\circ+30^\circ$ , $105^\circ=60^\circ+45^\circ$ .
✓
The tangent form follows by dividing $\sin(A\pm B)$ by $\cos(A\pm B)$ and then dividing numerator and denominator by $\cos A\cos B$ .

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Get a Paper 1 solution marked: find exact values with the double-angle identities

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Khan Academy: Inverse Trigonometric Functions↗

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS3.5 Trigonometric identities and equations NEXT →3.7 Vectors: properties and operations

Revise Compound and double angle identities; inverse trig for IB Mathematics: Analysis and Approaches (3.6)

In simple terms

Explore the concept

Key formulas

Full topic notes

The compound-angle (addition) identities

The double-angle identities

The inverse trigonometric functions

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 1 solution marked: find exact values with the double-angle identities

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

Full topic notes

The compound-angle (addition) identities

The double-angle identities

The inverse trigonometric functions

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 1 solution marked: find exact values with the double-angle identities

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Compound and double angle identities; inverse trig