MathsISC Class 11 · Compound and Multiple Angles

Transformation Formulae: Sum to Product, Product to Sum and Conditional Identities

Updated 17 October 2026 · Draft, not yet fact-checked

Sum-to-product: sinC + sinD = 2 sin((C + D)/2) cos((C − D)/2), and similar. Product-to-sum: 2 sinA cosB = sin(A + B) + sin(A − B). When A + B + C = π, results such as sin2A + sin2B + sin2C = 4 sinA sinB sinC and tanA + tanB + tanC = tanA tanB tanC hold.

Understand Transformation Formulae and Conditional Identities

Sum and difference to product. $sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2}$ ; $sin C - sin D = 2 cos \frac{C + D}{2} sin \frac{C - D}{2}$ ; $cos C + cos D = 2 cos \frac{C + D}{2} cos \frac{C - D}{2}$ ; $cos C - cos D = - 2 sin \frac{C + D}{2} sin \frac{C - D}{2}$ .

Product to sum. $2 sin A cos B = sin (A + B) + sin (A - B)$ ; $2 cos A sin B = sin (A + B) - sin (A - B)$ ; $2 cos A cos B = cos (A + B) + cos (A - B)$ ; $2 sin A sin B = cos (A - B) - cos (A + B)$ .

Special products. $cos A cos 2 A cos 4 A ... cos 2^{n - 1} A = \frac{s i n 2 ^{n} A}{2 ^{n} s i n A}$ , so $cos 20° cos 40° cos 80° = \frac{s i n 160°}{8 s i n 20°} = \frac{1}{8}$ . Also $sin 20° sin 40° sin 80° = \frac{3}{8}$ .

Conditional identities. If A + B + C = π (angles of a triangle): $sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C$ ; $cos 2 A + cos 2 B + cos 2 C = - 1 - 4 cos A cos B cos C$ ; $tan A + tan B + tan C = tan A tan B tan C$ ; $sin A + sin B + sin C = 4 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}$ . Use C = π − (A + B) to convert.

Strategy for proofs. Group terms into pairs that give a common factor, convert sums to products, take out the common factor and simplify, often ending with a product of three sines or cosines.

Transformation Formulae and Conditional Identities: formulas to remember

Sum to product: $\sin C + \sin D = 2\sin \frac{C+D}{2} \cos \frac{C-D}{2}$; For the sine difference: sin C − sin D = 2 cos((C+D)/2) sin((C−D)/2). Note that the sin and cos factors swap.
Cosine sum and difference: $\cos C + \cos D = 2\cos \frac{C+D}{2} \cos \frac{C-D}{2}$ | $\cos C - \cos D = -2\sin \frac{C+D}{2} \sin \frac{C-D}{2}$; Note the minus sign.
Product to sum: $2\sin A \cos B = \sin(A+B) + \sin(A-B)$; Also 2cosA cosB and 2sinA sinB.
Triangle identity: $If $A + B + C = \pi$: $\tan A + \tan B + \tan C = \tan A \tan B \tan C$$; And sin 2A + sin 2B + sin 2C = 4 sinA sinB sinC.

Transformation Formulae and Conditional Identities 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.

Beat pattern in sound

Suppose two tuning forks produce sounds $y = sin (5 t) + sin (3 t)$ (t in seconds, arbitrary units).

Write the combined sound as a product and identify the fast and slow factors.

Try it first, then show the solutionSolution

$sin 5 t + sin 3 t = 2 sin 4 t cos t$ .
The factor $sin 4 t$ is the fast oscillation (the average frequency).
The factor $cos t$ is the slow envelope that makes the loudness rise and fall (beats).

Answer: 2 sin 4t cos t: fast sin 4t with slow cos t envelope.

Memory hook: A sum of sines becomes a product of an average and a difference term.

Tricks to remember Transformation Formulae and Conditional Identities

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

Average and half-difference
Sum formulas use (C + D)/2 and (C − D)/2.
Use it when: Remembering sum to product.
Multiply and divide by sin A
To evaluate cosA cos2A cos4A, multiply and divide by 2 sinA to keep doubling the angle.
Use it when: Special products.

Transformation Formulae and Conditional Identities: solved examples

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

Example 1: Sum to product · 1 mark · MCQBoard level

$sin 75° + sin 15°$ equals:
(a) $\frac{6}{2}$
(b) $\frac{3}{2}$
(c) $\frac{1}{2}$
(d) $\frac{2}{2}$

Try it first, then show the solutionSolution

$2 sin 45° cos 30° = 2 \cdot \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{2} = \frac{6}{2}$ .

Answer: (a) $\frac{6}{2}$

Example 2: A continued product · 3 marks · Short answerBoard level

Prove that $cos 20° cos 40° cos 80° = \frac{1}{8}$ .

Try it first, then show the solutionSolution

Multiply and divide by $2 sin 20°$ : $\frac{2 s i n 20° c o s 20° c o s 40° c o s 80°}{2 s i n 20°} = \frac{s i n 40° c o s 40° c o s 80°}{2 s i n 20°}$ . (1 mark)
$= \frac{s i n 80° c o s 80°}{4 s i n 20°} = \frac{s i n 160°}{8 s i n 20°}$ . (1 mark)
$sin 160° = sin 20°$ , so the value is $\frac{1}{8}$ . (1 mark)

Answer: 1/8

Example 3: Quotient of sums · 4 marks · Short answerHarder board level

Prove that $\frac{s i n 5 x + s i n 3 x}{c o s 5 x + c o s 3 x} = tan 4 x$ . Hence find its value at x = 15°.

Try it first, then show the solutionSolution

Numerator $= 2 sin 4 x cos x$ . (1 mark)
Denominator $= 2 cos 4 x cos x$ . (1 mark)
Ratio $= \frac{s i n 4 x}{c o s 4 x} = tan 4 x$ . (1 mark)
At x = 15°: $tan 60° = 3$ . (1 mark)

Answer: Proved; value √3.

Example 4: A product of four sines · 4 marks · Short answerHarder board level

Find the value of $sin 20° sin 40° sin 60° sin 80°$ .

Try it first, then show the solutionSolution

$sin 60° = \frac{3}{2}$ . (0.5 mark)
Use $sin A sin (60° - A) sin (60° + A) = \frac{1}{4} sin 3 A$ with A = 20°: $sin 20° sin 40° sin 80° = \frac{1}{4} sin 60° = \frac{3}{8}$ . (2.5 marks)
Product $= \frac{3}{8} \times \frac{3}{2} = \frac{3}{16}$ . (1 mark)

Answer: 3/16

Example 5: A conditional identity · 5 marks · Long answerHOTS / competency

If A + B + C = π, prove that $sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C$ . Verify with A = B = C = 60°.

Try it first, then show the solutionSolution

$sin 2 A + sin 2 B = 2 sin (A + B) cos (A - B)$ . (1 mark)
Since $A + B = π - C$ , $sin (A + B) = sin C$ , and $sin 2 C = 2 sin C cos C = - 2 sin C cos (A + B)$ because $cos C = - cos (A + B)$ . (1.5 marks)
Sum $= 2 sin C [cos (A - B) - cos (A + B)] = 2 sin C \cdot 2 sin A sin B = 4 sin A sin B sin C$ . (1.5 marks)
Check: LHS $= 3 sin 120° = \frac{3 3}{2}$ ; RHS $= 4 (\frac{3}{2})^{3} = \frac{3 3}{2}$ . (1 mark)

Answer: Proved; both sides equal 3√3/2 at 60°, 60°, 60°.

Where students lose marks in Transformation Formulae and Conditional Identities

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

Mistake: Forgetting the minus sign in cos C − cos D
Writing +2 sin sin.
Do this: cos C − cos D = −2 sin((C + D)/2) sin((C − D)/2).
Usually costs: 2 marks.
Mistake: Wrong halves
Using (C + D) instead of (C + D)/2.
Do this: Half the sum and half the difference.
Usually costs: 1 mark.
Mistake: Not using C = π − (A + B)
Leaving the conditional identity unproved.
Do this: Substitute for one angle in terms of the others.
Usually costs: 1 to 2 marks.

Exam tips

Pair the terms to get a common factor.
Write the formula you use at each step.
Check results with specific angles.
In conditional identities, use A + B = π − C early.

Practice questions on Transformation Formulae and Conditional Identities

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
$2 sin 3 x cos x$ equals:
(a) $sin 4 x + sin 2 x$
(b) $sin 4 x - sin 2 x$
(c) $cos 4 x + cos 2 x$
(d) $cos 2 x - cos 4 x$
Hint
Product to sum.
Answer
(a) $sin 4 x + sin 2 x$
Q2 · 2 marks · Short answerBoard level
Express $cos 7 x + cos 3 x$ as a product.
Hint
Use the cosine sum formula.
Answer
$2 cos 5 x cos 2 x$ .
Q3 · 3 marks · Short answerHarder board level
Prove that $cos 10° - sin 10° = 2 sin 35°$ numerically to three decimals and by identity.
Hint
Write cos 10° as sin 80°.
Answer
$sin 80° - sin 10° = 2 cos 45° sin 35° = 2 sin 35°$ . Numerically $0.9848 - 0.1736 = 0.8112$ and $1.4142 \times 0.5736 = 0.8112$ .
Q4 · 3 marks · Short answerHarder board level
If A + B + C = π, show that $tan A + tan B + tan C = tan A tan B tan C$ and check for A = 45°, B = 60°.
Hint
C = 75°.
Answer
$tan (A + B) = - tan C$ gives the result. A = 45°, B = 60°, C = 75°: LHS $= 1 + 1.7321 + 3.7321 = 6.4642$ ; RHS $= 1 \times 1.7321 \times 3.7321 = 6.4641$ .
Q5 · 3 marks · Short answerHarder board level
Evaluate $sin 105° + sin 15°$ exactly.
Hint
Use the sum formula.
Answer
$2 sin 60° cos 45° = 2 \cdot \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{2} = \frac{6}{2}$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose two radio signals of the form $sin (7 t)$ and $sin (5 t)$ are added.
(i) Write their sum as a product. (1 mark)
(ii) Name the fast and slow factors. (1 mark)
(iii) Find the sum at $t = \frac{π}{12}$ exactly. (1 mark)
(iv) Find the value of t in $(0, π /2)$ where the slow factor is zero. (1 mark)
Hint
Average 6t, half-difference t.
Answer
(i) $2 sin 6 t cos t$ . (ii) Fast: $sin 6 t$ ; slow: $cos t$ . (iii) $2 sin (\frac{π}{2}) cos (\frac{π}{12}) = 2 cos 15° = \frac{6 + 2}{2} \approx 1.9319$ . (iv) $cos t = 0$ at $t = π /2$ , the end of the interval, so there is none strictly inside.

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 C + sin D?Show answer
2 sin((C+D)/2) cos((C−D)/2).
2.cos 20° cos 40° cos 80°?Show answer
1/8.
3.tan A + tan B + tan C when A+B+C=π?Show answer
tan A tan B tan C.
4.2 sin A cos B?Show answer
sin(A+B) + sin(A−B).

Transformation Formulae and Conditional Identities: frequently asked questions

When should I use product-to-sum?

When the expression is a product of sines or cosines that needs to become a sum, for example before integrating or comparing.

Why do we multiply and divide by sin A in special products?

Each step of 2 sin θ cos θ = sin 2θ doubles the angle and builds a telescoping product.

Do conditional identities hold if A + B + C ≠ π?

No. They depend on the condition A + B + C = π (for example the angles of a triangle).