MathsISC Class 11 · Chapter 4

Compound and Multiple Angles ISC Class 11: Formulae and Proofs

Q: Which formulae must I be able to derive for ISC Class 11?

The syllabus asks you to express sin(x ± y) and cos(x ± y) and to deduce tan(x ± y), cot(x ± y), the sum-product identities and the double and triple angle identities. Practise deriving each from the addition formulae.

Q: Are half-angle and one-third-angle formulae included?

Yes. The syllabus lists double, triple, half and one-third angle formulae as special cases of the addition formulae.

Q: How do I decide the sign in a half-angle formula?

Find the interval in which x/2 lies (halve the given interval for x), then use the quadrant signs for x/2.

Q: Why are sum-to-product formulae useful?

A product can be cancelled or factorised, a sum cannot. Turning sin A + sin B into 2 sin(...) cos(...) often creates a common factor in a fraction, simplifying it at once.

Q: How can I check a formula I am unsure of?

Substitute simple angles such as x = 30° and y = 60° (or x = 0) into both sides. If the two sides differ, the formula is wrong.

Q: What does tan(A + B + C) give when A, B, C are the angles of a triangle?

Since A + B + C = π and tan π = 0, the numerator must vanish: tan A + tan B + tan C = tan A tan B tan C.

Updated 15 October 2026 · Fact-checked

Compound and Multiple Angles is a chapter in Unit 1 of ISC Class 11 Mathematics. It covers sin(x ± y), cos(x ± y), tan(x ± y), cot(x ± y) and tan(A + B + C), double, triple, half and one-third angle formulae, and converting sums into products and products into sums, used for proofs and exact values.

What Compound and Multiple Angles is about

Knowing sin x and sin y does not tell you sin(x + y) by simple addition: sin(30° + 60°) = 1, but sin 30° + sin 60° ≈ 1.37. This chapter gives the correct rules for compound angles (sums and differences) and multiple angles (2x, 3x, x/2, x/3).

Addition formulae. $cos (x \pm y) = cos x cos y \mp sin x sin y$ and $sin (x \pm y) = sin x cos y \pm cos x sin y$ . Dividing gives $tan (x \pm y) = \frac{t a n x \pm t a n y}{1 \mp t a n x t a n y}$ and $cot (x \pm y) = \frac{c o t x c o t y \mp 1}{c o t y \pm c o t x}$ , and extending to three angles gives tan(A + B + C). These give exact values such as sin 75° and tan 15°.

Multiple angles. Putting y = x: sin 2x = 2 sin x cos x; cos 2x = cos²x − sin²x = 2cos²x − 1 = 1 − 2sin²x; tan 2x = 2 tan x/(1 − tan²x). Triple angles: sin 3x = 3 sin x − 4 sin³x; cos 3x = 4 cos³x − 3 cos x; tan 3x = (3 tan x − tan³x)/(1 − 3 tan²x). Replacing x by x/2 or x/3 gives the half-angle and one-third-angle forms.

Sums and products. $sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2}$ , and three companions, turn sums into products; reading them backwards, 2 sin A cos B = sin(A + B) + sin(A − B) and its companions turn products into sums. These are the main tools for proving identities and simplifying expressions.

Weightage: Compound and Multiple Angles belongs to the unit Sets and Functions, which CISCE weights at 18 of the 80 marks in the ISC Class XI Mathematics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Mathematics syllabus, examination year 2028, Class XI unit weightage

Key concepts

sin(x ± y) and cos(x ± y): cos(x + y) = cos x cos y − sin x sin y and cos(x − y) = cos x cos y + sin x sin y. These can be proved by comparing two expressions for the distance between points on the unit circle. Replacing x by π/2 − x gives sin(x + y) = sin x cos y + cos x sin y and sin(x − y) = sin x cos y − cos x sin y. Note the signs: in the cos formulae the sign in the middle is opposite to that in the bracket. Useful consequences: sin(x + y) sin(x − y) = sin²x − sin²y and cos(x + y) cos(x − y) = cos²x − sin²y. The formulae give exact values for 15°, 75°, 105° and so on.; Worked example
Find the exact values of sin 75° and cos 75°.
Try it first, then show the solutionSolution
sin 75° = sin(45° + 30°) = (1/√2)(√3/2) + (1/√2)(1/2) = (√3 + 1)/(2√2) = (√6 + √2)/4
cos 75° = cos(45° + 30°) = (1/√2)(√3/2) − (1/√2)(1/2) = (√6 − √2)/4
Answer: (√6 + √2)/4 ≈ 0.966; (√6 − √2)/4 ≈ 0.259
tan(x ± y), cot(x ± y) and tan(A + B + C): Dividing the sin formula by the cos formula and then dividing top and bottom by cos x cos y: tan(x ± y) = (tan x ± tan y)/(1 ∓ tan x tan y). Similarly cot(x ± y) = (cot x cot y ∓ 1)/(cot y ± cot x). Special cases: tan(π/4 + x) = (1 + tan x)/(1 − tan x) and tan(π/4 − x) = (1 − tan x)/(1 + tan x). For three angles: tan(A + B + C) = (tan A + tan B + tan C − tan A tan B tan C)/(1 − tan A tan B − tan B tan C − tan C tan A). If A + B + C = π, the numerator must be zero, so tan A + tan B + tan C = tan A tan B tan C.; Worked example
Find tan 15° exactly, and show that if A + B = 45°, then (1 + tan A)(1 + tan B) = 2.
Try it first, then show the solutionSolution
tan 15° = tan(45° − 30°) = (1 − 1/√3)/(1 + 1/√3) = (√3 − 1)/(√3 + 1) = 2 − √3
tan(A + B) = 1, so tan A + tan B = 1 − tan A tan B, i.e. tan A + tan B + tan A tan B = 1
(1 + tan A)(1 + tan B) = 1 + (tan A + tan B + tan A tan B) = 2
Answer: tan 15° = 2 − √3; the product is 2
Double-angle formulae: Putting y = x in the addition formulae: sin 2x = 2 sin x cos x = 2 tan x/(1 + tan²x); cos 2x = cos²x − sin²x = 2cos²x − 1 = 1 − 2sin²x = (1 − tan²x)/(1 + tan²x); tan 2x = 2 tan x/(1 − tan²x). Rearranged, they give the 'power-reducing' forms 2cos²x = 1 + cos 2x and 2sin²x = 1 − cos 2x, which are used constantly in proofs. The tan forms let you write sin 2x and cos 2x from tan x alone.; Worked example
If tan x = 3/4 and x is acute, find sin 2x, cos 2x and tan 2x.
Try it first, then show the solutionSolution
sin 2x = 2(3/4)/(1 + 9/16) = (3/2)/(25/16) = 24/25
cos 2x = (1 − 9/16)/(1 + 9/16) = 7/25
tan 2x = 24/7
Answer: 24/25; 7/25; 24/7
Triple-angle formulae: Writing 3x = 2x + x and using the double-angle results: sin 3x = 3 sin x − 4 sin³x; cos 3x = 4 cos³x − 3 cos x; tan 3x = (3 tan x − tan³x)/(1 − 3 tan²x). Rearranged: 4 sin³x = 3 sin x − sin 3x and 4 cos³x = 3 cos x + cos 3x. Notice the pattern of signs: sin 3x starts with 3 sin x, cos 3x starts with 4 cos³x.; Worked example
Prove that sin 3x = 3 sin x − 4 sin³x and check it for x = 30°.
Try it first, then show the solutionSolution
Write 3x as x + 2x: sin 3x = sin x cos 2x + cos x sin 2x = sin x(1 − 2sin²x) + cos x(2 sin x cos x)
= sin x − 2 sin³x + 2 sin x(1 − sin²x) = 3 sin x − 4 sin³x
Check: x = 30°: 3(½) − 4(⅛) = 1.5 − 0.5 = 1 = sin 90° ✓
Answer: Proved; checks for x = 30°
Half-angle and one-third-angle formulae: Replacing x by x/2 in the double-angle formulae: sin x = 2 sin(x/2) cos(x/2); cos x = 2cos²(x/2) − 1 = 1 − 2sin²(x/2); tan x = 2 tan(x/2)/(1 − tan²(x/2)). Hence sin(x/2) = ±√((1 − cos x)/2), cos(x/2) = ±√((1 + cos x)/2) and tan(x/2) = (1 − cos x)/sin x = sin x/(1 + cos x). The sign depends on the quadrant of x/2, not of x. Replacing x by x/3 in the triple-angle formulae gives the one-third-angle forms, e.g. sin x = 3 sin(x/3) − 4 sin³(x/3).; Worked example
If cos x = −1/3 and π < x < 3π/2, find sin(x/2), cos(x/2) and tan(x/2).
Try it first, then show the solutionSolution
π/2 < x/2 < 3π/4, so x/2 is in Q II: sin positive, cos negative.
sin(x/2) = √((1 + 1/3)/2) = √(2/3); cos(x/2) = −√((1 − 1/3)/2) = −√(1/3)
tan(x/2) = √(2/3)/(−√(1/3)) = −√2
Answer: √(2/3); −1/√3; −√2
Sum and difference to product: Adding and subtracting the addition formulae and putting x + y = C, x − y = D: sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2); sin C − sin D = 2 cos((C + D)/2) sin((C − D)/2); cos C + cos D = 2 cos((C + D)/2) cos((C − D)/2); cos C − cos D = −2 sin((C + D)/2) sin((C − D)/2). The minus sign in the last one is easy to forget. These are used to simplify fractions such as (sin 7x + sin 5x)/(cos 7x + cos 5x) and to solve or prove identities.; Worked example
Prove that (sin 7x + sin 5x)/(cos 7x + cos 5x) = tan 6x.
Try it first, then show the solutionSolution
Numerator = 2 sin 6x cos x; denominator = 2 cos 6x cos x
Ratio = sin 6x/cos 6x = tan 6x
Answer: Proved
Product to sum or difference: Reading the addition formulae the other way: 2 sin A cos B = sin(A + B) + sin(A − B); 2 cos A sin B = sin(A + B) − sin(A − B); 2 cos A cos B = cos(A + B) + cos(A − B); 2 sin A sin B = cos(A − B) − cos(A + B). These turn products into sums, which are easier to add, compare or (later) integrate. A product of three cosines can be reduced by applying the rule twice.; Worked example
Evaluate 2 sin 75° cos 15° and 2 sin 75° sin 15°.
Try it first, then show the solutionSolution
2 sin 75° cos 15° = sin 90° + sin 60° = 1 + √3/2
2 sin 75° sin 15° = cos 60° − cos 90° = ½
Answer: 1 + √3/2; ½

Maths in real life: Compound and Multiple Angles

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.

Adding two slopes

Suppose a hiking path climbs gently at an angle α with tan α = 1/7 and then turns up more steeply by a further angle β with tan β = 3/4 (both measured in the same vertical plane).

What total angle does the second part of the path make with the horizontal?

Try it first, then show the solutionSolution

tan(α + β) = (1/7 + 3/4)/(1 − (1/7)(3/4)) = (25/28)/(25/28) = 1
α + β = 45°

Answer: 45°, even though neither angle alone is a 'nice' angle.

Memory hook: Add tangents through the tan(A + B) formula, never directly.

Beats from sum to product

Imagine two tuning forks sounding together, their combined sound being y = sin(880πt) + sin(888πt).

Write y as a product and find how many times a second the loudness rises and falls.

Try it first, then show the solutionSolution

sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2) = 2 sin(884πt) cos(4πt)
The slowly varying amplitude 2 cos(4πt) peaks in size twice per cycle; cos(4πt) has frequency 2 Hz, so loudness peaks 4 times a second

Answer: y = 2 sin(884πt) cos(4πt): a 442 Hz note whose loudness pulses 4 times a second (beats).

Memory hook: Sum to product turns two close notes into one note with a slow pulse.

Double angle in projectile range

Suppose a ball is thrown at 20 m s⁻¹ and its range on level ground is R = (u²/g) sin 2θ, with g = 10 m s⁻².

Compare the ranges at θ = 30° and θ = 60°, and find the greatest range.

Try it first, then show the solutionSolution

θ = 30°: sin 60° = √3/2, R = 40 × 0.866 = 34.6 m; θ = 60°: sin 120° = √3/2, the same range
Greatest when sin 2θ = 1, θ = 45°: R = 40 m

Answer: 34.6 m for both angles; 40 m at 45°.

Memory hook: sin 2θ is symmetric about 45°.

Topics in this chapter

Compound and Multiple Angles formula sheet

Addition formulae (sin, cos): $sin (x \pm y) = sin x cos y \pm cos x sin y; cos (x \pm y) = cos x cos y \mp sin x sin y$; Opposite sign in the cos formula.
Addition formulae (tan, cot): $tan (x \pm y) = \frac{t a n x \pm t a n y}{1 \mp t a n x t a n y}; cot (x \pm y) = \frac{c o t x c o t y \mp 1}{c o t y \pm c o t x}$; tan(π/4 + x) = (1 + tan x)/(1 − tan x).
tan of three angles: $tan (A + B + C) = \frac{\sum t a n A - t a n A t a n B t a n C}{1 - \sum t a n A t a n B}$; A + B + C = π ⇒ Σ tan A = Π tan A.
Double angles: $sin 2 x = 2 sin x cos x; cos 2 x = 2 cos^{2} x - 1 = 1 - 2 sin^{2} x; tan 2 x = \frac{2 t a n x}{1 - t a n ^{2} x}$; sin 2x = 2tan x/(1 + tan²x).
Triple angles: $sin 3 x = 3 sin x - 4 sin^{3} x; cos 3 x = 4 cos^{3} x - 3 cos x; tan 3 x = \frac{3 t a n x - t a n ^{3} x}{1 - 3 t a n ^{2} x}$; Check with x = 30°.
Half angles: $sin \frac{x}{2} = \pm \frac{1 - c o s x}{2}; cos \frac{x}{2} = \pm \frac{1 + c o s x}{2}; tan \frac{x}{2} = \frac{1 - c o s x}{s i n x}$; Sign from the quadrant of x/2.
Sum to product: $sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2}; cos C - cos D = - 2 sin \frac{C + D}{2} sin \frac{C - D}{2}$; Also sin C − sin D and cos C + cos D.
Product to sum: $2 sin A cos B = sin (A + B) + sin (A - B); 2 sin A sin B = cos (A - B) - cos (A + B)$; Also 2cos A sin B and 2cos A cos B.

Compound and Multiple Angles: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Unit-circle picture for cos(x − y)

Proving the addition formula for cos(x − y).

Draw

A unit circle centred at O with points P(cos x, sin x) and Q(cos y, sin y), so that angle POQ = x − y.
A second copy: the same angle x − y placed from the x-axis, with points A(1, 0) and R(cos(x − y), sin(x − y)).
Join PQ and AR, which are equal chords because they subtend equal angles.

Label

O
P(cos x, sin x)
Q(cos y, sin y)
A(1, 0)
R(cos(x − y), sin(x − y))
x − y

Watch out: Equate PQ² and AR² using the distance formula; the squares of sin and cos add to 1 and leave cos(x − y) = cos x cos y + sin x sin y.

Sign chart for half angles

Choosing the sign in half-angle formulae.

Draw

Two number lines: one for x from 0 to 2π (or the given range), one for x/2 directly beneath it.
Mark the quadrant boundaries π/2, π, 3π/2 on each.
Shade the given interval of x and its image for x/2, then read off the quadrant of x/2.

Label

x
x/2
π/2
π
3π/2
quadrant of x/2

Watch out: The sign of sin(x/2) and cos(x/2) depends on the quadrant of x/2, not of x.

How the school exam asks Compound and Multiple Angles

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Exact values	Find sin 15°, cos 75°, tan 105°, tan 22.5° and similar.	Write the angle as a sum or difference (or half) of standard angles.
Given ratios, find another	Given sin x and cos y in stated quadrants, find sin(x + y) or tan(x − y).	Find the missing ratios with correct signs first, then substitute.
Multiple and half angles	Find sin 2x, cos 3x or sin(x/2) from a given ratio; prove triple-angle formulae.	Use the tan forms or the power-reducing forms; fix the sign by the quadrant of the new angle.
Sum-product conversions	Simplify (sin A + sin B)/(cos A + cos B); express products as sums; prove identities.	Pair terms so that (C + D)/2 or (C − D)/2 gives a common factor.
Conditional identities	If A + B + C = π, prove tan A + tan B + tan C = tan A tan B tan C, or similar.	Write C = π − (A + B) and use tan(π − θ) = −tan θ.

Tricks to remember Compound and Multiple Angles

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

cos is contrary
In cos(x ± y) the middle sign is the opposite of the bracket sign; in sin(x ± y) it is the same.
Use it when: Writing the addition formulae.
Sin-Cos-Cos-Sin
sin(x + y) = sin cos + cos sin: 'SC + CS'; cos(x + y) = cos cos − sin sin: 'CC − SS'.
Use it when: Remembering the order of terms.
Three, four, four, three
sin 3x = 3 sin x − 4 sin³x; cos 3x = 4 cos³x − 3 cos x.
Use it when: Triple angles.
Half-angle sign
Look at where x/2 lives, not x.
Use it when: Choosing ± in half-angle formulae.
cos minus cos is minus
cos C − cos D carries a leading minus: −2 sin sin.
Use it when: Sum-to-product conversions.

Compound and Multiple Angles: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: A form of cos 2x · 1 mark · MCQExam level

cos 2x is equal to:
(a) 2 sin x cos x
(b) 1 − 2 sin²x
(c) 1 + 2 sin²x
(d) 2 cos²x + 1

Try it first, then show the solutionSolution

cos 2x = cos²x − sin²x = (1 − sin²x) − sin²x = 1 − 2 sin²x.

Answer: (b)

Example 2: Doubling an angle does not double the sine · 1 mark · Assertion-reasonExam level

Assertion: sin 2x = 2 sin x for all real x.
Reason: sin 2x = 2 sin x cos x.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

sin 2x = 2 sin x only when cos x = 1 or sin x = 0; e.g. x = 30° gives sin 60° ≈ 0.87, not 1. The Reason is the correct formula.

Answer: (d)

Example 3: Exact values for 165° · 2 marks · Short answerHarder exam level

Find the exact value of cos 165° and of tan 165°. [2]

Try it first, then show the solutionSolution

cos 165° = cos(120° + 45°) = cos 120° cos 45° − sin 120° sin 45° = (−½)(1/√2) − (√3/2)(1/√2) = −(1 + √3)/(2√2) = −(√6 + √2)/4
tan 165° = tan(120° + 45°) = (−√3 + 1)/(1 − (−√3)(1)) = (1 − √3)/(1 + √3) = −(2 − √3) = √3 − 2

Answer: −(√6 + √2)/4 ≈ −0.966; √3 − 2 ≈ −0.268

Example 4: A fourth-power identity · 3 marks · Short answerHarder exam level

Prove that cos 4x = 1 − 8 sin²x cos²x. [3]

Try it first, then show the solutionSolution

cos 4x = 1 − 2 sin²2x (double-angle formula with 2x)
sin 2x = 2 sin x cos x, so sin²2x = 4 sin²x cos²x
cos 4x = 1 − 8 sin²x cos²x

Answer: Proved

Example 5: Sum-to-product simplification · 3 marks · Short answerHarder exam level

Prove that (sin 5x + sin 3x)/(sin 5x − sin 3x) = tan 4x cot x. [3]

Try it first, then show the solutionSolution

Numerator: sin 5x + sin 3x = 2 sin 4x cos x
Denominator: sin 5x − sin 3x = 2 cos 4x sin x
Ratio = (sin 4x/cos 4x)(cos x/sin x) = tan 4x cot x

Answer: Proved

Example 6: Triple and half angles · 5 marks · Long answerHarder exam level

(a) Prove that tan 3x = (3 tan x − tan³x)/(1 − 3 tan²x). [2]
(b) If tan x = −4/3 and π/2 < x < π, find sin(x/2), cos(x/2) and tan(x/2). [2]
(c) Find the exact value of tan 22.5°. [1]

Try it first, then show the solutionSolution

(a) tan 3x = tan(2x + x) = (tan 2x + tan x)/(1 − tan 2x tan x), with tan 2x = 2t/(1 − t²), t = tan x.
= [2t + t(1 − t²)]/[(1 − t²) − 2t²] = (3t − t³)/(1 − 3t²)
(b) In Q II, cos x = −3/5. x/2 lies between π/4 and π/2 (Q I), so both are positive: sin(x/2) = √((1 + 3/5)/2) = √(4/5) = 2/√5
cos(x/2) = √((1 − 3/5)/2) = √(1/5) = 1/√5; tan(x/2) = 2
(c) tan 22.5° = (1 − cos 45°)/sin 45° = (1 − 1/√2)/(1/√2) = √2 − 1

Answer: (b) 2/√5, 1/√5, 2 (c) √2 − 1

Example 7: Angles of a triangle · 3 marks · Short answerHOTS / competency

If A + B + C = π, prove that tan A + tan B + tan C = tan A tan B tan C. [3]

Try it first, then show the solutionSolution

A + B = π − C, so tan(A + B) = tan(π − C) = −tan C
(tan A + tan B)/(1 − tan A tan B) = −tan C
tan A + tan B = −tan C + tan A tan B tan C, so tan A + tan B + tan C = tan A tan B tan C

Answer: Proved

Example 8: Product of three cosines · 3 marks · Short answerHOTS / competency

Prove that cos 20° cos 40° cos 80° = 1/8. [3]

Try it first, then show the solutionSolution

Multiply and divide by 2 sin 20°: 2 sin 20° cos 20° = sin 40°
Then 2 sin 40° cos 40° = sin 80°, and 2 sin 80° cos 80° = sin 160° = sin 20°
So 8 sin 20° × (product) = sin 20°, giving product = 1/8

Answer: Proved (= 0.125)

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Angles of elevation from one point

Suppose from a point P on level ground, the top of a 10 m pole is seen at an angle of elevation α with tan α = ½, and the top of a taller tower directly behind it is seen at an angle α + β, where tan β = ⅓.

(i) · 1 mark
Find the distance of P from the foot of the pole.
Answer
Answer: 20 m
tan α = 10/d = ½.
(ii) · 1 mark
Find tan(α + β).
Answer
Answer: 1
(½ + ⅓)/(1 − 1/6) = 1.
(iii) · 1 mark
What is the angle of elevation of the tower's top?
Answer
Answer: 45°
tan(α + β) = 1.
(iv) · 2 marks
If the tower is 30 m from P, find its height, and find sin 2α.
Answer
Answer: 30 m; 4/5
Height = 30 tan 45° = 30 m. sin 2α = 2 tan α/(1 + tan²α) = 1/(5/4) = 4/5.

Combining two waves

Suppose two signals are added: y = sin 5t + sin 3t.

(i) · 1 mark
Write y as a product.
Answer
Answer: y = 2 sin 4t cos t
sin C + sin D with C = 5t, D = 3t.
(ii) · 1 mark
Find y when t = π/8.
Answer
Answer: 2 cos(π/8) ≈ 1.85
sin(π/2) = 1, so y = 2 cos(π/8) ≈ 1.848.
(iii) · 1 mark
Find all t in [0, π] where y = 0.
Answer
Answer: t = 0, π/4, π/2, 3π/4, π
sin 4t = 0 gives multiples of π/4; cos t = 0 gives π/2 (already included).
(iv) · 2 marks
Express sin 5t sin 3t as a difference of cosines and use it to find its value at t = π/6.
Answer
Answer: ½(cos 2t − cos 8t); ½ at t = π/6
2 sin A sin B = cos(A − B) − cos(A + B), so sin 5t sin 3t = ½(cos 2t − cos 8t). At t = π/6: ½(cos(π/3) − cos(4π/3)) = ½(½ + ½) = ½ (check: sin(5π/6) sin(π/2) = ½ × 1 = ½).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the exact value of sin 15°. (2 marks)

Step	Marks
sin 15° = sin(45° − 30°) = sin 45° cos 30° − cos 45° sin 30°	1
= (√3 − 1)/(2√2) = (√6 − √2)/4	1

If sin x = 3/5 and cos y = −12/13, with x in Q II and y in Q III, find sin(x + y). (3 marks)

Step	Marks
x in Q II: cos x = −4/5	1
y in Q III: sin y = −5/13	1
sin(x + y) = (3/5)(−12/13) + (−4/5)(−5/13) = −36/65 + 20/65 = −16/65	1

Prove that (sin 5x − 2 sin 3x + sin x)/(cos 5x − cos x) = tan x. (5 marks)

Step	Marks
sin 5x + sin x = 2 sin 3x cos 2x	1
Numerator = 2 sin 3x cos 2x − 2 sin 3x = 2 sin 3x (cos 2x − 1) = −4 sin 3x sin²x	1
cos 5x − cos x = −2 sin 3x sin 2x	1
Ratio = (−4 sin 3x sin²x)/(−2 sin 3x sin 2x) = 2 sin²x/sin 2x	1
= 2 sin²x/(2 sin x cos x) = tan x	1

Where students lose marks in Compound and Multiple Angles

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

Mistake: Treating sin as distributive
Students write sin(x + y) = sin x + sin y or sin 2x = 2 sin x.
Wrong working
sin 75° = sin 45° + sin 30° = 1.21.
Right working
sin 75° = sin 45° cos 30° + cos 45° sin 30° = 0.966.
Do this: Check any 'formula' with x = 30°, y = 60° before using it.
Usually costs: Full marks for the part.
Mistake: Wrong sign in cos(x + y)
Students copy the + sign from inside the bracket.
Wrong working
cos(x + y) = cos x cos y + sin x sin y.
Right working
cos(x + y) = cos x cos y − sin x sin y.
Do this: 'cos is contrary': opposite sign in the middle.
Usually costs: 1-2 marks.
Mistake: Choosing the half-angle sign from the quadrant of x
Students look at x instead of x/2.
Wrong working
x in Q III, so cos(x/2) is negative and sin(x/2) negative.
Right working
For π < x < 3π/2, x/2 is in Q II: sin(x/2) > 0, cos(x/2) < 0.
Do this: Halve the interval first.
Usually costs: 1 mark.
Mistake: Losing the minus sign in cos C − cos D
It is the only sum-product formula with a leading minus.
Wrong working
cos 5x − cos x = 2 sin 3x sin 2x.
Right working
cos 5x − cos x = −2 sin 3x sin 2x.
Do this: Check with C = 0, D = π: cos 0 − cos π = 2, and −2 sin(π/2) sin(−π/2) = 2 ✓.
Usually costs: 1 mark.
Mistake: Using the wrong signs for the given quadrants
In 'find sin(x + y)' problems, students take all ratios as positive.
Wrong working
cos x = 4/5 for x in Q II.
Right working
cos x = −4/5 in Q II.
Do this: Write the quadrant signs (ASTC) before substituting.
Usually costs: 1 mark per value.

Practice questions on Compound and Multiple Angles

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 · MCQExam level
sin 2x is equal to:
(a) 2 sin x
(b) 2 sin x cos x
(c) sin²x − cos²x
(d) 1 − 2 sin²x
Hint
Double angle.
Answer
(b)
Q2 · 1 mark · MCQExam level
The value of sin 75° cos 15° + cos 75° sin 15° is:
(a) 0
(b) ½
(c) √3/2
(d) 1
Hint
sin(A + B).
Answer
(d) sin 90° = 1
Q3 · 1 mark · Very short answerExam level
Write the formula for cos 3x in terms of cos x.
Hint
Starts with 4.
Answer
cos 3x = 4 cos³x − 3 cos x
Q4 · 1 mark · Very short answerExam level
Express 2 cos 5x cos 3x as a sum.
Hint
2 cos A cos B.
Answer
cos 8x + cos 2x
Q5 · 2 marks · Short answerExam level
Find the exact value of tan 105°.
Hint
60° + 45°.
Answer
(√3 + 1)/(1 − √3) = −(2 + √3)
Q6 · 2 marks · Short answerHarder exam level
If tan A = 2 and tan B = 3, with A and B acute, find A + B.
Hint
tan(A + B), then decide the quadrant.
Answer
tan(A + B) = (2 + 3)/(1 − 6) = −1. Both angles exceed 45° (tan > 1), so A + B lies between π/2 and π: A + B = 3π/4.
Q7 · 2 marks · Short answerHarder exam level
If sin x = 4/5 with x acute, find sin 3x.
Hint
3 sin x − 4 sin³x.
Answer
12/5 − 4(64/125) = 300/125 − 256/125 = 44/125
Q8 · 2 marks · Short answerHarder exam level
Prove that (1 − cos 2x)/sin 2x = tan x.
Hint
1 − cos 2x = 2 sin²x.
Answer
2 sin²x/(2 sin x cos x) = sin x/cos x = tan x.
Q9 · 2 marks · Short answerHarder exam level
Show that sin(x + y) sin(x − y) = sin²x − sin²y.
Hint
Expand both brackets.
Answer
(sin x cos y)² − (cos x sin y)² = sin²x(1 − sin²y) − (1 − sin²x)sin²y = sin²x − sin²y.
Q10 · 3 marks · Short answerHarder exam level
Prove that cos 80° + cos 40° − cos 20° = 0.
Hint
Combine the first two.
Answer
cos 80° + cos 40° = 2 cos 60° cos 20° = cos 20°, so the expression is cos 20° − cos 20° = 0.
Q11 · 3 marks · Short answerHarder exam level
If cos x = 4/5 and 3π/2 < x < 2π, find sin(x/2), cos(x/2) and tan(x/2).
Hint
x/2 is in Q II.
Answer
3π/4 < x/2 < π: sin(x/2) = √((1 − 4/5)/2) = 1/√10; cos(x/2) = −√((1 + 4/5)/2) = −3/√10; tan(x/2) = −1/3.
Q12 · 5 marks · Long answerHarder exam level
(a) Prove that sin 3x = 3 sin x − 4 sin³x. [2]
(b) Prove that sin x sin(60° − x) sin(60° + x) = ¼ sin 3x. [2]
(c) Hence find sin 20° sin 40° sin 80°. [1]
Hint
sin(60° − x) sin(60° + x) = sin²60° − sin²x.
Answer
(a) sin(2x + x) = 2 sin x cos²x + sin x − 2 sin³x = 3 sin x − 4 sin³x. (b) sin x(3/4 − sin²x) = ¼(3 sin x − 4 sin³x) = ¼ sin 3x. (c) x = 20°: ¼ sin 60° = √3/8.
Q13 · 3 marks · Short answerHOTS / competency
If A + B + C = π, prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.
Hint
Combine the first two by sum to product.
Answer
sin 2A + sin 2B = 2 sin(A + B) cos(A − B) = 2 sin C cos(A − B); sin 2C = 2 sin C cos C = −2 sin C cos(A + B). Sum = 2 sin C[cos(A − B) − cos(A + B)] = 2 sin C × 2 sin A sin B = 4 sin A sin B sin C.

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.cos(x − y)?Show answer
cos x cos y + sin x sin y.
2.tan(x + y)?Show answer
(tan x + tan y)/(1 − tan x tan y).
3.cos 2x in terms of sin x?Show answer
1 − 2 sin²x.
4.sin 3x?Show answer
3 sin x − 4 sin³x.
5.tan(x/2) without square roots?Show answer
(1 − cos x)/sin x.
6.sin C − sin D?Show answer
2 cos((C + D)/2) sin((C − D)/2).
7.2 sin A sin B?Show answer
cos(A − B) − cos(A + B).
8.tan 15°?Show answer
2 − √3.

Best order to study it

Learn and prove cos(x − y), then derive cos(x + y), sin(x ± y).
Derive tan(x ± y), cot(x ± y) and tan(A + B + C); practise exact values.
Derive the double-angle formulae and their tan forms.
Derive the triple-angle formulae; then half-angle and one-third-angle forms.
Learn the four sum-to-product formulae and the four product-to-sum formulae.
Practise proving identities, including conditional ones with A + B + C = π.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write all four addition formulae with correct signs?
Can you find sin 75°, cos 15° and tan 105° exactly?
Can you find sin(x + y) given ratios in different quadrants?
Can you write cos 2x in three ways?
Can you prove sin 3x and tan 3x formulae?
Can you find half-angle values with correct signs?
Can you convert sums to products and back?
Can you prove tan A + tan B + tan C = tan A tan B tan C when A + B + C = π?

Compound and Multiple Angles: frequently asked questions

Which formulae must I be able to derive for ISC Class 11?

The syllabus asks you to express sin(x ± y) and cos(x ± y) and to deduce tan(x ± y), cot(x ± y), the sum-product identities and the double and triple angle identities. Practise deriving each from the addition formulae.

Are half-angle and one-third-angle formulae included?

Yes. The syllabus lists double, triple, half and one-third angle formulae as special cases of the addition formulae.

How do I decide the sign in a half-angle formula?

Find the interval in which x/2 lies (halve the given interval for x), then use the quadrant signs for x/2.

Why are sum-to-product formulae useful?

A product can be cancelled or factorised, a sum cannot. Turning sin A + sin B into 2 sin(...) cos(...) often creates a common factor in a fraction, simplifying it at once.

How can I check a formula I am unsure of?

Substitute simple angles such as x = 30° and y = 60° (or x = 0) into both sides. If the two sides differ, the formula is wrong.

What does tan(A + B + C) give when A, B, C are the angles of a triangle?

Since A + B + C = π and tan π = 0, the numerator must vanish: tan A + tan B + tan C = tan A tan B tan C.

Based on: CISCE ISC 2028 Mathematics syllabus (Classes XI and XII); CISCE Regulations for the ISC Examination Year 2028 (Class XI internal examination)