MathsISC Class 11 · Compound and Multiple Angles

Compound, Double and Triple Angle Formulae: Exact Values and Identities

Updated 17 October 2026 · Fact-checked

The key results are sin(A ± B) = sinA cosB ± cosA sinB, cos(A ± B) = cosA cosB ∓ sinA sinB and tan(A ± B) = (tanA ± tanB)/(1 ∓ tanA tanB). From these, sin2A = 2sinA cosA, cos2A = cos²A − sin²A and sin3A = 3sinA − 4sin³A.

Every multiple-angle formula comes from sin(A+B): set B = A for double angles, then use 2A + A to reach triple angles.

Illustration drawn for QuestionBuddy.

Understand Sum, Difference, Double and Triple Angle Formulae

Compound angles. $sin (A \pm B) = sin A cos B \pm cos A sin B$ ; $cos (A \pm B) = cos A cos B \mp sin A sin B$ ; $tan (A \pm B) = \frac{t a n A \pm t a n B}{1 \mp t a n A t a n B}$ . For three angles, $tan (A + B + C) = \frac{t a n A + t a n B + t a n C - t a n A t a n B t a n C}{1 - t a n A t a n B - t a n B t a n C - t a n C t a n A}$ .

Exact values. Use 15° = 45° − 30° and 75° = 45° + 30°: $sin 75° = \frac{6 + 2}{4}$ , $cos 75° = \frac{6 - 2}{4}$ , $tan 75° = 2 + 3$ , $tan 15° = 2 - 3$ .

Double angles. $sin 2 A = 2 sin A cos A = \frac{2 t a n A}{1 + t a n ^{2} A}$ ; $cos 2 A = cos^{2} A - sin^{2} A = 2 cos^{2} A - 1 = 1 - 2 sin^{2} A = \frac{1 - t a n ^{2} A}{1 + t a n ^{2} A}$ ; $tan 2 A = \frac{2 t a n A}{1 - t a n ^{2} A}$ . The half-angle forms: $2 sin^{2} A = 1 - cos 2 A$ and $2 cos^{2} A = 1 + cos 2 A$ .

Triple angles. $sin 3 A = 3 sin A - 4 sin^{3} A$ ; $cos 3 A = 4 cos^{3} A - 3 cos A$ ; $tan 3 A = \frac{3 t a n A - t a n ^{3} A}{1 - 3 t a n ^{2} A}$ .

Strategy. When the value of one ratio of A is given, find the others from the identity and the quadrant (including sign) before applying the formula. In identities, replace multiple angles by single-angle functions (or the reverse) and simplify.

Sum, Difference, Double and Triple Angle Formulae: formulas to remember

Compound angles: $\sin(A ± B) = \sin A \cos B ± \cos A \sin B$ | $\cos(A ± B) = \cos A \cos B ∓ \sin A \sin B$; Signs of cos reverse.
Tangent: $\tan(A ± B) = \frac{\tan A ± \tan B}{1 ∓ \tan A \tan B}$; Valid when the denominator ≠ 0.
Double angle: $\sin 2A = 2\sin A \cos A$ | $\cos 2A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$ | $\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$; cos 2A has three forms.
Triple angle: $\sin 3A = 3\sin A - 4\sin^3 A$ | $\cos 3A = 4\cos^3 A - 3\cos A$; From A + 2A.

Sum, Difference, Double and Triple Angle Formulae in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A ramp of 15 degrees

Suppose an engineer needs the exact slope ratio of a ramp inclined at 15° (as 45° − 30°).

Find sin 15° and tan 15° exactly.

Try it first, then show the solutionSolution

$sin 15° = sin (45° - 30°) = \frac{1}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot \frac{1}{2} = \frac{3 - 1}{2 2} = \frac{6 - 2}{4}$ .
$tan 15° = \frac{1 - 1/ 3}{1 + 1/ 3} = \frac{3 - 1}{3 + 1} = 2 - 3$ .
Numerically $sin 15° = 0.2588$ and $tan 15° = 0.2679$ .

Answer: sin 15° = (√6 − √2)/4 ≈ 0.2588; tan 15° = 2 − √3 ≈ 0.2679.

Memory hook: Express the angle as a sum or difference of 30°, 45°, 60°.

Tricks to remember Sum, Difference, Double and Triple Angle Formulae

Short memory aids. Say each one aloud once, then use it on a question the same day.

cos flips the sign
In cos(A + B) the sign between the products is minus; in cos(A − B) it is plus.
Use it when: Compound angle formulae.
Three forms of cos 2A
Choose cos²A − sin²A, 2cos²A − 1 or 1 − 2sin²A to match the given data.
Use it when: Simplifying and proving.

Sum, Difference, Double and Triple Angle Formulae: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Exact value · 1 mark · MCQBoard level

The value of $cos 15°$ is:
(a) $\frac{6 - 2}{4}$
(b) $\frac{6 + 2}{4}$
(c) $\frac{3 + 1}{2}$
(d) $\frac{3 - 1}{2}$

Try it first, then show the solutionSolution

$cos (45° - 30°) = \frac{1}{2} \cdot \frac{3}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{6 + 2}{4}$ .

Answer: (b) $\frac{6 + 2}{4}$

Example 2: Values of 75° · 3 marks · Short answerBoard level

Find the exact values of $cos 75°$ and $cot 75°$ .

Try it first, then show the solutionSolution

$cos 75° = cos (45° + 30°) = cos 45° cos 30° - sin 45° sin 30°$ . (1 mark)
$= \frac{1}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot \frac{1}{2} = \frac{3 - 1}{2 2} = \frac{6 - 2}{4}$ . (1 mark)
$tan 75° = \frac{1 + 1/ 3}{1 - 1/ 3} = \frac{( 3 + 1 ) ^{2}}{2} = 2 + 3$ , so $cot 75° = \frac{1}{2 + 3} = 2 - 3$ . (1 mark)

Answer: cos 75° = (√6 − √2)/4; cot 75° = 2 − √3.

Example 3: sin(A + B) from given ratios · 4 marks · Short answerHarder board level

If $sin A = \frac{3}{5}$ (A acute) and $cos B = - \frac{12}{13}$ (B in the second quadrant), find $sin (A + B)$ and $cos (A + B)$ .

Try it first, then show the solutionSolution

$cos A = \frac{4}{5}$ and $sin B = \frac{5}{13}$ (positive in the second quadrant). (1 mark)
$sin (A + B) = \frac{3}{5} \cdot (- \frac{12}{13}) + \frac{4}{5} \cdot \frac{5}{13} = \frac{- 36 + 20}{65} = - \frac{16}{65}$ . (1.5 marks)
$cos (A + B) = \frac{4}{5} \cdot (- \frac{12}{13}) - \frac{3}{5} \cdot \frac{5}{13} = \frac{- 48 - 15}{65} = - \frac{63}{65}$ . (1.5 marks)

Answer: sin(A + B) = −16/65; cos(A + B) = −63/65.

Example 4: Double angle from tan A · 4 marks · Short answerHarder board level

If $tan A = \frac{3}{4}$ , find $sin 2 A$ , $cos 2 A$ and $tan 2 A$ , and verify that $sin^{2} 2 A + cos^{2} 2 A = 1$ .

Try it first, then show the solutionSolution

$sin 2 A = \frac{2 \cdot 3/4}{1 + 9/16} = \frac{3/2}{25/16} = \frac{24}{25}$ . (1.5 marks)
$cos 2 A = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$ . (1 mark)
$tan 2 A = \frac{24}{7}$ . (0.5 mark)
Check: $(24/25)^{2} + (7/25)^{2} = (576 + 49) /625 = 1$ . (1 mark)

Answer: sin 2A = 24/25; cos 2A = 7/25; tan 2A = 24/7.

Example 5: Triple angle identity · 5 marks · Long answerHOTS / competency

Prove that $\frac{s i n 3 A}{s i n A} - \frac{c o s 3 A}{c o s A} = 2$ . Hence find the value of $sin 3 A$ if $sin A = \frac{1}{3}$ .

Try it first, then show the solutionSolution

$sin 3 A = 3 sin A - 4 sin^{3} A$ gives $\frac{s i n 3 A}{s i n A} = 3 - 4 sin^{2} A$ . (1.5 marks)
$cos 3 A = 4 cos^{3} A - 3 cos A$ gives $\frac{c o s 3 A}{c o s A} = 4 cos^{2} A - 3$ . (1.5 marks)
Difference $= 3 - 4 sin^{2} A - 4 cos^{2} A + 3 = 6 - 4 = 2$ . (1 mark)
For $sin A = \frac{1}{3}$ : $sin 3 A = 3 \cdot \frac{1}{3} - 4 \cdot \frac{1}{27} = 1 - \frac{4}{27} = \frac{23}{27}$ . (1 mark)

Answer: Proved; sin 3A = 23/27.

Where students lose marks in Sum, Difference, Double and Triple Angle Formulae

The slips examiners see most often, and what to write instead.

Mistake: Wrong sign in cos(A + B)
Writing cosA cosB + sinA sinB.
Do this: cos(A + B) has a minus sign.
Usually costs: 2 marks.
Mistake: Ignoring the quadrant
Taking sin positive in the third quadrant.
Do this: Use ASTC to find the sign.
Usually costs: 1 to 2 marks.
Mistake: Rationalising incorrectly
Leaving (√3 + 1)/(√3 − 1) unsimplified.
Do this: Multiply by the conjugate: result 2 + √3.
Usually costs: 1 mark.

Exam tips

Write the formula before substituting.
Find all required ratios and their signs first.
Check numerically with a calculator where possible.
For multiple angles, express in terms of the single angle.

Practice questions on Sum, Difference, Double and Triple Angle Formulae

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
$cos (A + B) + cos (A - B)$ equals:
(a) $2 sin A sin B$
(b) $2 cos A cos B$
(c) $2 sin A cos B$
(d) $- 2 sin A sin B$
Hint
Add the two expansions.
Answer
(b) $2 cos A cos B$
Q2 · 2 marks · Short answerBoard level
Find the value of $sin 105°$ .
Hint
Use 60° + 45°.
Answer
$sin 105° = sin 60° cos 45° + cos 60° sin 45° = \frac{3 + 1}{2 2} = \frac{6 + 2}{4}$ .
Q3 · 3 marks · Short answerHarder board level
Prove that $\frac{1 - c o s 2 A}{1 + c o s 2 A} = tan^{2} A$ .
Hint
Use the half-angle forms.
Answer
$1 - cos 2 A = 2 sin^{2} A$ and $1 + cos 2 A = 2 cos^{2} A$ , so the ratio is $tan^{2} A$ .
Q4 · 3 marks · Short answerHarder board level
If $cos A = - \frac{3}{5}$ and A is in the third quadrant, find $sin 2 A$ and $cos 2 A$ .
Hint
Find sin A with the right sign.
Answer
$sin A = - \frac{4}{5}$ ; $sin 2 A = 2 \cdot (- \frac{4}{5}) (- \frac{3}{5}) = \frac{24}{25}$ ; $cos 2 A = 2 \cdot \frac{9}{25} - 1 = - \frac{7}{25}$ .
Q5 · 3 marks · Short answerHarder board level
Show that $tan A + tan (60° + A) + tan (120° + A) = 3 tan 3 A$ for A = 10° numerically to three decimals.
Hint
Use tan 10°, tan 70°, tan 130°.
Answer
$tan 10° = 0.1763$ , $tan 70° = 2.7475$ , $tan 130° = - 1.1918$ ; sum $= 1.7320$ . $3 tan 30° = 3 \times 0.5774 = 1.7321$ . They agree.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a surveyor measures two angles of elevation A and B of a tower from two points, with $tan A = \frac{1}{2}$ and $tan B = \frac{1}{3}$ .
(i) Find $tan (A + B)$ . (1 mark)
(ii) State the value of A + B. (1 mark)
(iii) Find $tan (A - B)$ . (1 mark)
(iv) Find $tan 2 A$ . (1 mark)
Hint
Use the tangent formulae.
Answer
(i) $\frac{1/2 + 1/3}{1 - 1/6} = \frac{5/6}{5/6} = 1$ . (ii) A + B = 45°. (iii) $\frac{1/2 - 1/3}{1 + 1/6} = \frac{1/6}{7/6} = \frac{1}{7}$ . (iv) $\frac{2 \cdot 1/2}{1 - 1/4} = \frac{4}{3}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.sin 2A?Show answer
2 sin A cos A.
2.cos 3A?Show answer
4cos³A − 3cos A.
3.tan 75°?Show answer
2 + √3.
4.cos(A + B)?Show answer
cosA cosB − sinA sinB.

Sum, Difference, Double and Triple Angle Formulae: frequently asked questions

Why do we have three forms of cos 2A?

They are equal through sin²A + cos²A = 1; the best form depends on whether the sine or cosine (or tangent) is given.

How do I get sin 3A?

Write 3A = 2A + A, apply the sum formula and the double angle formulae.

Are these formulae true for all angles?

Yes (except where tangents are undefined).