3.10 Compound angle identities · Educerie Mathematics: analysis and approaches

This whole subtopic is higher level. Nothing in it is on an SL paper.

Educerie · IB Diploma · Mathematics: analysis and approaches

Topic 3 Geometry and trigonometry · 3.10 Compound angle identities

Level: HL only, the whole subtopic. If you are SL, none of this is on your papers.
Themes (key concepts): equivalence, generalization, relationships. A compound angle identity is an equivalence between the ratio of a sum and the ratios of its parts, and the double angle identities you met at SL turn out to be one special case of this generalization.
The question this unit answers: how do you find sin, cos or tan of a sum or difference of two angles from the ratios of the angles themselves, and where do the double angle identities come from?
Where it is examined: Paper 1 (no calculator): exact values such as sin 75° or tan(π/12) (3 to 4 marks), sin(A + B) given sin A and cos B (4 to 6 marks), "show that" identities, and equations that open up once a compound angle is expanded (5 to 7 marks). Paper 2: the same equations with non-exact solutions. Paper 3: identities such as sin 3θ built step by step. The identities are in the formula booklet; the marks are for choosing and using them.

What you must be able to do

You must be able to	Level	What it looks like in the exam
Use sin(A ± B), cos(A ± B) and tan(A ± B) to find exact values	HL only	"Find the exact value of cos(7π/12)" (3 marks)
Find the ratio of a sum or difference from given ratios of A and B, with signs set by the quadrant	HL only	"Given sin A = 3/5 … find sin(A + B)" (4 to 5 marks)
Recognise an expanded compound angle and collapse it	HL only	"Write cos 50° cos 10° − sin 50° sin 10° as a single trigonometric ratio" (2 marks)
Derive the double angle identities for sin, cos and tan from the compound angle identities	HL only	"Show that tan 2θ = 2 tan θ / (1 − tan²θ)" (3 marks)
Use tan 2θ = 2 tan θ / (1 − tan²θ) to find values and solve equations	HL only	"Given tan 2x = 3/4 and x acute, find tan x" (5 marks)
Prove identities and solve equations using compound and double angles, including triple angles	HL only	"Show that cos 3θ = 4cos³θ − 3cos θ. Hence solve …" (4 + 3 marks)

Before you start

You need the unit circle and exact values (SL 3.5), the identity cos²θ + sin²θ = 1 and the double angle identities for sin and cos (SL 3.6), and how to solve a trigonometric equation in an interval (SL 3.8). The proof in section 3 uses the distance formula (SL 3.1) and the cosine rule (SL 3.2). From 3.9 you need cot, sec and cosec. Radians unless the question uses degrees.

1The idea in one paragraph

sin(A + B) is not sin A + sin B. Try A = B = π/4: sin(π/2) = 1, but sin(π/4) + sin(π/4) = √2 ≈ 1.41. The ratio of a sum mixes the ratios of both parts in a fixed pattern, and that pattern is the compound angle identity: sin(A + B) = sin A cos B + cos A sin B, with partners for cos and tan and for differences. Those six formulas do three jobs. They give exact values of angles like 15° and 75° from the ones you know. They find sin(A + B) when you know only sin A and cos B. And they generate every double angle identity: put B = A and sin(A + A) becomes 2 sin A cos A. The identities are in the booklet; the skill is recognising which one a question is hiding.

2The six identities

sin(A ± B) = sin A cos B ± cos A sin B cos(A ± B) = cos A cos B ∓ sin A sin B tan(A ± B) = (tan A ± tan B) ÷ (1 ∓ tan A tan B) · all in the formula booklet.

Read the signs carefully, because that is where the marks go.

Sine keeps the sign. sin(A + B) has a + in the middle; sin(A − B) has a −. Sine also mixes: sin cos + cos sin.
Cosine flips the sign. cos(A + B) has a − in the middle; cos(A − B) has a +. Cosine does not mix: cos cos, then sin sin.
Tangent keeps the sign on top and flips it underneath. tan(A + B) = (tan A + tan B) ÷ (1 − tan A tan B).

The ∓ symbol means "the opposite of the ± sign above": take the top sign everywhere or the bottom sign everywhere.

The tangent identity is not new information: it is the sine identity divided by the cosine identity. Divide the top and bottom of (sin A cos B + cos A sin B) ÷ (cos A cos B − sin A sin B) by cos A cos B and every term becomes a tangent or a 1. That is worth knowing because it shows why the tangent formula fails when tan A or tan B is undefined; use sine and cosine instead.

3Why they are true

The guide does not ask you to prove the compound angle identities, but seeing one proof makes the sign pattern stick, and the method reappears with vectors in 3.13.

Put two points on the unit circle, P at angle A and Q at angle B, as in Figure 1. Their coordinates are P(cos A, sin A) and Q(cos B, sin B), and the angle between OP and OQ is A − B. Now measure PQ² in two ways.

Figure 1 · Two points on the unit circle, two ways to measure PQ

PQ² = (cos A − cos B)² + (sin A − sin B)²distance formula

= cos² A + sin² A + cos² B + sin² B − 2cos A cos B − 2sin A sin B

= 2 − 2(cos A cos B + sin A sin B)cos² + sin² = 1, twice

PQ² = 1² + 1² − 2(1)(1)cos(A − B)cosine rule in triangle OPQ

= 2 − 2cos(A − B)

cos(A − B) = cos A cos B + sin A sin Bequate the two

Every other identity follows from this one. Replace B with −B, and use cos(−B) = cos B and sin(−B) = −sin B, to get cos(A + B) = cos A cos B − sin A sin B. For sine, use the fact that sin θ = cos(π/2 − θ):

sin(A + B) = cos(π/2 − A − B) = cos((π/2 − A) − B)

= cos(π/2 − A)cos B + sin(π/2 − A)sin B

= sin A cos B + cos A sin B

Figure 2 shows the addition formulas as a picture, which the guide encourages: stack a right-angled triangle with angle B and hypotenuse 1 on top of the line at angle A. The height of the top corner splits into sin A cos B plus cos A sin B, and its distance across is cos A cos B minus the overlap sin A sin B.

Figure 2 · sin(A + B) and cos(A + B) read off one picture

4Exact values from angles you know

Any angle that is a sum or difference of 30°, 45°, 60° and 90° (π/6, π/4, π/3, π/2) has an exact value you can build.

Worked example 1 (Paper 1). Find the exact values of sin 75°, cos(7π/12) and tan 15°.

sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°

= (√2/2)(√3/2) + (√2/2)(1/2)

= (√6 + √2)/4

cos(7π/12) = cos(π/3 + π/4) = cos(π/3)cos(π/4) − sin(π/3)sin(π/4)

= (1/2)(√2/2) − (√3/2)(√2/2)

= (√2 − √6)/4negative: 7π/12 is in the second quadrant

tan 15° = tan(45° − 30°) = (1 − 1/√3) / (1 + 1/√3)

= (√3 − 1)/(√3 + 1)multiply top and bottom by √3

= (√3 − 1)² / (3 − 1)rationalise

= (4 − 2√3)/2 = 2 − √3

A sign check costs nothing: sin 75° ≈ (2.449 + 1.414)/4 ≈ 0.966, close to 1, as it should be for an angle near 90°; cos(7π/12) is negative, as it must be just past π/2.

5From the ratios of A and B to the ratios of A ± B

This is the most common exam form. You are given one ratio of each angle and its quadrant, and you need both sin and cos of each before any identity can be used.

Worked example 2 (Paper 1). A is acute with sin A = 3/5, and B is obtuse with cos B = −5/13. Find the exact values of sin(A + B), cos(A − B) and tan(A + B).

cos A = 4/5A acute: cos A > 0 (3-4-5 triangle)

sin B = 12/13B obtuse: sin B > 0 (5-12-13 triangle)

sin(A + B) = (3/5)(−5/13) + (4/5)(12/13) = −15/65 + 48/65 = 33/65

cos(A − B) = (4/5)(−5/13) + (3/5)(12/13) = −20/65 + 36/65 = 16/65

tan A = 3/4, tan B = −12/5

tan(A + B) = (3/4 − 12/5) / (1 − (3/4)(−12/5))

= (−33/20) / (56/20) = −33/56

Check the last one another way: cos(A + B) = −20/65 − 36/65 = −56/65, and sin(A + B) ÷ cos(A + B) = (33/65) ÷ (−56/65) = −33/56. When the two routes agree, the answer is right.

6Reading an identity backwards

Examiners like to hide a compound angle in its expanded form. Spot the pattern (cos cos − sin sin, sin cos − cos sin) and collapse it.

Worked example 3 (Paper 1). Write each as a single trigonometric ratio and evaluate where possible.

cos 50° cos 10° − sin 50° sin 10° = cos(50° + 10°) = cos 60° = 1/2

sin 3x cos x − cos 3x sin x = sin(3x − x) = sin 2x

(tan 70° − tan 25°) / (1 + tan 70° tan 25°) = tan(70° − 25°) = tan 45° = 1

In the first line, the minus between cos cos and sin sin says plus inside the bracket, because cosine flips. That is the sign most often lost.

7Double angle identities, derived

At SL you were given sin 2θ and cos 2θ. At HL you must be able to derive them, and tan 2θ with them, by putting B = A. Figure 3 is the whole family tree.

Figure 3 · Every double angle identity is a compound angle identity with B = A

sin 2A = sin(A + A) = sin A cos A + cos A sin A = 2sin A cos A

cos 2A = cos(A + A) = cos A cos A − sin A sin A = cos² A − sin² A

= cos² A − (1 − cos² A) = 2cos² A − 1replace sin² A

= (1 − sin² A) − sin² A = 1 − 2sin² Aor replace cos² A

tan 2A = tan(A + A) = (tan A + tan A) / (1 − tan A tan A)

= 2tan A / (1 − tan² A)

tan 2θ = 2 tan θ ÷ (1 − tan²θ) · in the formula booklet. It fails where tan θ = ±1, because then 2θ is an odd multiple of π/2 and tan 2θ is undefined.

The three forms of cos 2A each have a job: cos²A − sin²A when both appear, 2cos²A − 1 to make an equation all in cos, 1 − 2sin²A to make it all in sin. Rearranged, the last two give cos²A = (1 + cos 2A)/2 and sin²A = (1 − cos 2A)/2, which you will use in HL integration.

The misconception to kill now. sin 2x is not 2 sin x. Figure 4 draws both: sin 2x squeezes the graph horizontally and still peaks at 1, while 2 sin x stretches it vertically to a peak of 2. They are equal only where sin x(2cos x − 2) = 0, which is where sin x = 0 or cos x = 1.

Figure 4 · sin 2x and 2 sin x are different functions

Worked example 4 (Paper 1). θ is acute and tan θ = 1/2. Find tan 2θ, sin 2θ and cos 2θ.

tan 2θ = 2(1/2) / (1 − 1/4) = 1 / (3/4) = 4/3

sin θ = 1/√5, cos θ = 2/√5right-angled triangle with sides 1, 2, √5

sin 2θ = 2(1/√5)(2/√5) = 4/5

cos 2θ = 4/5 − 1/5 = 3/5cos² θ − sin² θ

check: (4/5) ÷ (3/5) = 4/3 ✓

Another route, for HL (link to 1.14). De Moivre's theorem gives (cos θ + i sin θ)² = cos 2θ + i sin 2θ. Expanding the left side gives cos²θ − sin²θ + 2i sin θ cos θ. Matching real and imaginary parts produces both double angle identities at once. The same trick with the cube gives the triple angle identities, and a Paper 3 question may ask you to compare the two methods.

8Building bigger identities: triple angles

Write 3θ as 2θ + θ, expand with the compound identity, then use the double angle identities. The skill is choosing the form of cos 2θ that leaves the variable you want.

Worked example 5 (Paper 1). Show that sin 3θ = 3 sin θ − 4 sin³θ.

sin 3θ = sin(2θ + θ) = sin 2θ cos θ + cos 2θ sin θ

= (2sin θ cos θ)cos θ + (1 − 2sin² θ)sin θthe sin-only form of cos 2θ

= 2sin θ cos² θ + sin θ − 2sin³ θ

= 2sin θ(1 − sin² θ) + sin θ − 2sin³ θcos² θ = 1 − sin² θ

= 3sin θ − 4sin³ θ

"Show that" means every line must be visible. Jumping from line 1 to the answer scores at most the first method mark.

9Equations that open up

Expanding a compound angle often turns an awkward equation into a familiar one.

Worked example 6 (Paper 1). Solve sin(x + π/6) = 2 cos x for 0 ≤ x < 2π.

sin x cos(π/6) + cos x sin(π/6) = 2cos x

(√3/2)sin x + (1/2)cos x = 2cos x

(√3/2)sin x = (3/2)cos x

tan x = (3/2) ÷ (√3/2) = √3divide by cos x; cos x = 0 is not a solution

x = π/3, 4π/3tan has period π

Dividing by cos x is safe only after checking cos x = 0 does not solve the equation. Here, cos x = 0 would force sin(x + π/6) = 0, which does not happen at x = π/2 or 3π/2.

Worked example 7 (Paper 1). Solve tan 2x = 3 tan x for 0 ≤ x ≤ π.

2tan x / (1 − tan² x) = 3tan x

2tan x = 3tan x(1 − tan² x)

tan x(3tan² x − 1) = 0do not divide by tan x: that loses roots

tan x = 0 or tan x = ±1/√3

x = 0, π or x = π/6, 5π/6

Figure 5 confirms the four solutions. The graphs seem to cross at x = π/2, but 3 tan x is undefined there, so π/2 is not a solution. Moving everything to one side and factorising, instead of dividing by tan x, is what keeps x = 0 and x = π.

Figure 5 · Solutions of tan 2x = 3 tan x on 0 ≤ x ≤ π

On Paper 2 the same algebra often ends with a non-exact angle. Then solve for the whole compound angle first, list every value in the widened interval, and only then subtract. Q5 below works that way.

10Where marks are lost

Distributing the ratio over the sum. sin(A + B) ≠ sin A + sin B and cos 2x ≠ 2 cos x. Expand with the identity every time.

Getting the cosine sign backwards. cos(A + B) has a minus in the middle and cos(A − B) a plus. Test with A = B: cos(A − A) must be 1 = cos²A + sin²A.

Forgetting the quadrant when finding the missing ratios. If B is obtuse, cos B < 0 but sin B > 0. Using sin B = −12/13 ruins every later line.

Pairing tan with the wrong denominator. tan(A + B) has 1 − tan A tan B underneath; tan(A − B) has 1 + tan A tan B.

Dividing by tan x or sin x in an equation. It deletes the solutions where that factor is zero. Factorise instead.

Using the wrong form of cos 2θ. In an equation with sin θ, use 1 − 2sin²θ; with cos θ, use 2cos²θ − 1. The mixed form cos²θ − sin²θ leaves two variables.

Skipping lines in a "show that". The answer is printed, so the marks are all in the steps. Each substitution, and each use of cos²θ + sin²θ = 1, deserves its own line.

11Work it right

Before expanding, write the identity from the booklet with A and B named: "A = π/3, B = π/4".
For a ratio of A ± B from given data, find all four of sin A, cos A, sin B, cos B first, each with its sign and the quadrant that fixed it.
Bracket every substituted value, especially negatives: (4/5)(−5/13).
Check an exact value by estimating it, or by finding the same quantity a second way (sin ÷ cos = tan).
In a derivation, say "put B = A" and show the substitution before simplifying.
In an equation, collect everything on one side and factorise; never divide by an expression that could be zero. Check the interval and exclude values where a term is undefined.
For a compound argument like x + π/3 on Paper 2, widen the interval first, find every value, then subtract.

12Try it

Marks in brackets. Q1 to Q4 and Q6 are Paper 1 style (no calculator). Q5 is Paper 2 style.

Q1. Find the exact value of sin(π/12). 3 marks

Q2. Angles A and B are acute, with tan A = 1/3 and tan B = 1/2.

(a) Find the value of tan(A + B). 3 marks

(b) Hence find the exact value of A + B, justifying your answer. 2 marks

Q3. Show that sin 2θ ÷ (1 + cos 2θ) ≡ tan θ. 4 marks

Q4.

(a) Show that cos 3θ = 4cos³θ − 3cos θ. 4 marks

(b) Hence solve 8cos³θ − 6cos θ = 1 for 0 ≤ θ ≤ π. 3 marks

Q5. Paper 2.

(a) Show that 2 sin(x + π/3) = sin x + √3 cos x. 2 marks

(b) Hence solve sin x + √3 cos x = 1.5 for 0 ≤ x ≤ 2π. 4 marks

Q6. Given that tan 2x = 3/4 and 0 < x < π/2, find the exact value of tan x. 5 marks

13In one breath

The ratio of a sum is not the sum of the ratios: sin(A ± B) = sin A cos B ± cos A sin B keeps the sign and mixes, cos(A ± B) = cos A cos B ∓ sin A sin B flips the sign and does not mix, and tan(A ± B) = (tan A ± tan B) ÷ (1 ∓ tan A tan B) keeps it on top and flips it below. They come from measuring one chord of the unit circle two ways. Use them to build exact values from 30°, 45° and 60°, to find sin(A + B) once you have all four ratios with quadrant signs, and to collapse an expanded expression back into one ratio. Put B = A and they give sin 2A = 2 sin A cos A, cos 2A in its three forms, and tan 2A = 2 tan A ÷ (1 − tan²A); write 3θ = 2θ + θ for triple angles. In equations, expand, collect on one side and factorise, never dividing by something that might be zero.

Answers

Q1. sin(π/12) = sin(π/3 − π/4) = sin(π/3)cos(π/4) − cos(π/3)sin(π/4) = (√3/2)(√2/2) − (1/2)(√2/2) = (√6 − √2)/4. M1 for writing π/12 as a difference (or sum) of known angles and using the identity, A1 for correct exact values substituted, A1 for (√6 − √2)/4. π/4 − π/6 is equally good.

Q2. (a) tan(A + B) = (1/3 + 1/2) ÷ (1 − (1/3)(1/2)) = (5/6) ÷ (5/6) = 1. (b) A and B are both acute, so 0 < A + B < π, and the only angle in that interval with tangent 1 is A + B = π/4. (c) sin A = 1/√10, cos A = 3/√10, sin B = 1/√5, cos B = 2/√5. sin(A − B) = (1/√10)(2/√5) − (3/√10)(1/√5) = (2 − 3)/√50 = −1/(5√2) = −√2/10. (a) M1 for the tan(A + B) identity, A1 for 5/6 ÷ 5/6, A1 for 1. (b) R1 for the interval 0 < A + B < π, A1 for π/4. (c) M1 for finding sin and cos of both angles, M1 for the sin(A − B) identity, A1 for −√2/10 or equivalent. A positive answer scores M1 M1 A0.

Q3. LHS = 2 sin θ cos θ ÷ (1 + 2cos²θ − 1) = 2 sin θ cos θ ÷ 2cos²θ = sin θ ÷ cos θ = tan θ = RHS. M1 for sin 2θ = 2 sin θ cos θ, M1 for choosing cos 2θ = 2cos²θ − 1, A1 for the denominator 2cos²θ, A1 for cancelling to tan θ. Using cos²θ − sin²θ can still succeed, but only with a further correct step.

Q4. (a) cos 3θ = cos(2θ + θ) = cos 2θ cos θ − sin 2θ sin θ = (2cos²θ − 1)cos θ − 2 sin²θ cos θ = 2cos³θ − cos θ − 2(1 − cos²θ)cos θ = 2cos³θ − cos θ − 2cos θ + 2cos³θ = 4cos³θ − 3cos θ. (b) 8cos³θ − 6cos θ = 2cos 3θ, so cos 3θ = 1/2. With 0 ≤ θ ≤ π, 0 ≤ 3θ ≤ 3π, so 3θ = π/3, 5π/3, 7π/3 and θ = π/9, 5π/9, 7π/9. (a) M1 for writing 3θ = 2θ + θ and expanding, M1 for substituting double angle identities, M1 for sin²θ = 1 − cos²θ, A1 for the printed result with all steps shown. (b) M1 for cos 3θ = 1/2, M1 for the interval 0 ≤ 3θ ≤ 3π, A1 for all three values. Only π/9 scores M1 M0 A0.

Q5. (a) 2 sin(x + π/3) = 2(sin x cos(π/3) + cos x sin(π/3)) = 2((1/2)sin x + (√3/2)cos x) = sin x + √3 cos x. (b) 2 sin(x + π/3) = 1.5, so sin(x + π/3) = 0.75. For 0 ≤ x ≤ 2π, π/3 ≤ x + π/3 ≤ 7π/3. arcsin 0.75 = 0.848…, which is below π/3 = 1.047…, so it is rejected; the values in the interval are π − 0.848… = 2.293… and 2π + 0.848… = 7.131…. Subtract π/3: x = 1.25 and x = 6.08 (3 s.f.). (a) M1 for expanding, A1 for the exact values giving the result. (b) M1 for sin(x + π/3) = 0.75, M1 for the widened interval or a GDC graph over 0 ≤ x ≤ 2π, A1 for 1.25, A1 for 6.08. Including x = 0.848 − 1.047 < 0 loses the last A1.

Q6. 2t ÷ (1 − t²) = 3/4 with t = tan x, so 8t = 3 − 3t² and 3t² + 8t − 3 = 0, giving (3t − 1)(t + 3) = 0 and t = 1/3 or t = −3. Since 0 < x < π/2, tan x > 0, so tan x = 1/3. M1 for the tan 2x identity, A1 for the quadratic 3t² + 8t − 3 = 0, M1 for solving it, A1 for both roots, R1 for rejecting −3 because x is acute.

Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section AHL 3.10 Compound angle identities and the double angle identity for tan. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

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