MathsCBSE Class 11 · Trigonometric Functions

Compound-Angle, Sum-to-Product and Multiple-Angle Identities with Proofs (Class 11)

Q: How do I choose between the three forms of cos 2x?

Pick the form that leaves the required function: use 1 − 2sin²x for sine, 2cos²x − 1 for cosine, and cos²x − sin²x for a difference of squares.

Q: Why does the sum-to-product formula use half angles?

Because the average of A and B and half of their difference appear when you add the expansions of sin(P + Q) and sin(P − Q) with P = (A + B)/2 and Q = (A − B)/2.

Q: Is the proof of the triple-angle formula needed?

Yes, deriving sin 3x and cos 3x from the sum formula is a standard board-level question.

Q: How can I check an identity quickly?

Put in a convenient angle such as 30° or 45° and compare the two sides numerically.

Updated 17 October 2026 · Fact-checked

Key identities: sin(A ± B) = sin A cos B ± cos A sin B and cos(A ± B) = cos A cos B ∓ sin A sin B; sin 2x = 2 sin x cos x, cos 2x = cos²x − sin²x = 2cos²x − 1 = 1 − 2sin²x; sin 3x = 3 sin x − 4 sin³x, cos 3x = 4cos³x − 3 cos x. Sum-to-product: sin A + sin B = 2 sin((A+B)/2) cos((A−B)/2). Also a sin x + b cos x has maximum √(a² + b²).

Put B = A in sin(A+B) to get sin 2x, then add x to get sin 3x = 3 sin x − 4 sin³x.

Illustration drawn for QuestionBuddy.

Understand Sum, Product and Multiple-Angle Identities with Proofs

Sum and difference formulae.
$ $sin (A \pm B) = sin A cos B \pm cos A sin B, cos (A \pm B) = cos A cos B \mp sin A sin B,$ $
$ $tan (A \pm B) = \frac{t a n A \pm t a n B}{1 \mp t a n A t a n B} .$ $
These give exact values at angles such as $7 5^{\circ} = 4 5^{\circ} + 3 0^{\circ}$ : $cos 7 5^{\circ} = \frac{6 - 2}{4}$ , $cos 1 5^{\circ} = \frac{6 + 2}{4}$ , and $tan 7 5^{\circ} = 2 + 3$ .

Multiple angles.

$sin 2 x = 2 sin x cos x = \frac{2 tan x}{1 + tan ^{2} x}$ .
$cos 2 x = cos^{2} x - sin^{2} x = 2 cos^{2} x - 1 = 1 - 2 sin^{2} x = \frac{1 - tan ^{2} x}{1 + tan ^{2} x}$ .
$sin 3 x = 3 sin x - 4 sin^{3} x$ , $cos 3 x = 4 cos^{3} x - 3 cos x$ , $tan 3 x = \frac{3 tan x - tan ^{3} x}{1 - 3 tan ^{2} x}$ .

**Proof of $sin 3 x$ :** $sin (2 x + x) = sin 2 x cos x + cos 2 x sin x = 2 sin x (1 - sin^{2} x) + (1 - 2 sin^{2} x) sin x = 3 sin x - 4 sin^{3} x$ .

Sum to product and product to sum.
$ $sin A + sin B = 2 sin \frac{A + B}{2} cos \frac{A - B}{2}, sin A - sin B = 2 cos \frac{A + B}{2} sin \frac{A - B}{2},$ $
$ $cos A + cos B = 2 cos \frac{A + B}{2} cos \frac{A - B}{2}, cos A - cos B = - 2 sin \frac{A + B}{2} sin \frac{A - B}{2} .$ $
Reversing: $2 sin A cos B = sin (A + B) + sin (A - B)$ , $2 cos A cos B = cos (A + B) + cos (A - B)$ , $2 sin A sin B = cos (A - B) - cos (A + B)$ .

Strategies for proofs.
1. Start with the more complicated side and aim for the other.
2. Group terms in pairs with the same sum (same half-sum), e.g. the first with the last and the two middle ones (as in $sin x + sin 3 x + sin 5 x + sin 7 x = (sin 7 x + sin x) + (sin 5 x + sin 3 x)$ ), and apply sum-to-product.
3. For products of sines or cosines with angles doubling, multiply and divide by $2 sin$ (or $sin$ of the smallest angle) and use $2 sin A cos A = sin 2 A$ repeatedly; e.g. $cos 2 0^{\circ} cos 4 0^{\circ} cos 8 0^{\circ} = \frac{1}{8}$ .
4. For products like $sin 2 0^{\circ} sin 4 0^{\circ} sin 8 0^{\circ}$ use $sin x sin (6 0^{\circ} - x) sin (6 0^{\circ} + x) = \frac{1}{4} sin 3 x$ .

**Extreme values of $a sin x + b cos x$ .** Write it as $R sin (x + α)$ with $R = a^{2} + b^{2}$ , $cos α = \frac{a}{R}$ , $sin α = \frac{b}{R}$ (from the sum formula). Then the maximum is $R$ and the minimum $- R$ . For $3 sin x + 4 cos x$ , $R = 5$ , the maximum $5$ occurs where $sin x = \frac{3}{5}$ and $cos x = \frac{4}{5}$ .

Sum, Product and Multiple-Angle Identities with Proofs: formulas to remember

Sine of a sum: $s in (A + B) = s in A cos B + cos A s in B$; Replace B by −B for the difference.
Cosine of a sum: $cos (A + B) = cos A cos B - s in A s in B$; Sign flips for the difference.
Double and triple angles: $s in 2 x = 2 s in x cos x; s in 3 x = 3 s in x - 4 s i n^{3} x; cos 3 x = 4 co s^{3} x - 3 cos x$; cos 2x = 2cos²x − 1 = 1 − 2sin²x.
Sum to product: $s in A + s in B = 2 s in ((A + B) /2) cos ((A - B) /2)$; Similar for the other three.
Extreme values: $a s in x + b cos x r an g eso v er [- \sqrt (a^{2} + b^{2}), \sqrt (a^{2} + b^{2})]$; Maximum at tan x = a/b when both positive.

How to solve Sum, Product and Multiple-Angle Identities with Proofs questions

Use these steps for identities, exact values and extreme values.

1Choose the side to simplify (usually the more complicated one) and decide the target.
2For exact values, split the angle into known angles ( $1 5^{\circ} = 4 5^{\circ} - 3 0^{\circ}$ , $7 5^{\circ} = 4 5^{\circ} + 3 0^{\circ}$ ) and use the sum or difference formula.
3For sums of sines or cosines, pair the terms and use sum-to-product, looking for a common factor.
4For products with doubling angles, multiply and divide by $2 sin$ of the smallest angle.
5For $a sin x + b cos x$ , write it as $R sin (x + α)$ and read off the maximum and minimum.
6Check the final answer numerically at one value.

Sum, Product and Multiple-Angle Identities with Proofs 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.

Height from an angle of elevation

Suppose a tower is seen from a point 40 m from its foot at an angle of elevation of $7 5^{\circ}$ .

Find the height using $tan 7 5^{\circ} = 2 + 3$ .

Try it first, then show the solutionSolution

$tan 7 5^{\circ} = tan (4 5^{\circ} + 3 0^{\circ}) = \frac{1 + \frac{1}{3}}{1 - \frac{1}{3}} = \frac{3 + 1}{3 - 1} = 2 + 3$ .
Height $= 40 (2 + 3) \approx 40 \times 3.732 = 149.3$ m.

Answer: About 149.3 m.

Memory hook: Build awkward angles from 30, 45 and 60.

Adding two tones

Suppose two sound waves have displacements $sin 5 0^{\circ}$ and $sin 1 0^{\circ}$ at a moment in time (in arbitrary units).

Find their sum without a calculator, as a single product.

Try it first, then show the solutionSolution

$sin 5 0^{\circ} + sin 1 0^{\circ} = 2 sin 3 0^{\circ} cos 2 0^{\circ}$ .
$= 2 \times \frac{1}{2} cos 2 0^{\circ} = cos 2 0^{\circ} \approx 0.940$ .

Answer: $cos 2 0^{\circ} \approx 0.940$

Memory hook: Sum of two sines becomes twice sine of the average times cosine of half the difference.

Tricks to remember Sum, Product and Multiple-Angle Identities with Proofs

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

Pair by equal sums
In a sum of four sines in A.P., pair the first with the last and the two middle ones.
Use it when: Sum-to-product proofs.
Double the angle trick
Multiply and divide by 2 sin of the smallest angle for products of doubling angles.
Use it when: cos 20° cos 40° cos 80°.
Three forms of cos 2x
Choose cos²x − sin²x, 2cos²x − 1 or 1 − 2sin²x to suit the target.
Use it when: Simplifying.
R sine alpha
a sin x + b cos x = R sin(x + α), R = √(a² + b²).
Use it when: Maxima and minima.

Sum, Product and Multiple-Angle Identities with Proofs: solved examples

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

Example 1: Exact value · 2 marks · Very short answerBoard level

Find the exact value of $cos 7 5^{\circ}$ .

Try it first, then show the solutionSolution

$cos 7 5^{\circ} = cos (4 5^{\circ} + 3 0^{\circ}) = cos 4 5^{\circ} cos 3 0^{\circ} - sin 4 5^{\circ} sin 3 0^{\circ}$ . (1 mark)
$= \frac{2}{2} \cdot \frac{3}{2} - \frac{2}{2} \cdot \frac{1}{2} = \frac{6 - 2}{4}$ . (1 mark)

Answer: $\frac{6 - 2}{4}$

Example 2: Proof of the triple-angle formula · 3 marks · Short answerBoard level

Prove that $sin 3 x = 3 sin x - 4 sin^{3} x$ .

Try it first, then show the solutionSolution

$sin 3 x = sin (2 x + x) = sin 2 x cos x + cos 2 x sin x$ . (1 mark)
$= 2 sin x cos^{2} x + (1 - 2 sin^{2} x) sin x = 2 sin x (1 - sin^{2} x) + sin x - 2 sin^{3} x$ . (1 mark)
$= 2 sin x - 2 sin^{3} x + sin x - 2 sin^{3} x = 3 sin x - 4 sin^{3} x$ . (1 mark)

Answer: Proved

Example 3: A sum of four sines · 3 marks · Short answerHarder board level

Prove that $sin x + sin 3 x + sin 5 x + sin 7 x = 4 cos x cos 2 x sin 4 x$ .

Try it first, then show the solutionSolution

Pair: $(sin 7 x + sin x) + (sin 5 x + sin 3 x)$ . (½ mark)
$sin 7 x + sin x = 2 sin 4 x cos 3 x$ and $sin 5 x + sin 3 x = 2 sin 4 x cos x$ . (1 mark)
Sum $= 2 sin 4 x (cos 3 x + cos x) = 2 sin 4 x \cdot 2 cos 2 x cos x$ . (1 mark)
$= 4 cos x cos 2 x sin 4 x$ . (½ mark)

Answer: Proved

Example 4: Product of sines · 3 marks · Short answerHarder board level

Find the value of $sin 2 0^{\circ} sin 4 0^{\circ} sin 6 0^{\circ} sin 8 0^{\circ}$ .

Try it first, then show the solutionSolution

Use $sin x sin (6 0^{\circ} - x) sin (6 0^{\circ} + x) = \frac{1}{4} sin 3 x$ with $x = 2 0^{\circ}$ : $sin 2 0^{\circ} sin 4 0^{\circ} sin 8 0^{\circ} = \frac{1}{4} sin 6 0^{\circ} = \frac{3}{8}$ . (2 marks)
Multiply by $sin 6 0^{\circ} = \frac{3}{2}$ : $\frac{3}{8} \cdot \frac{3}{2} = \frac{3}{16}$ . (1 mark)

Answer: $\frac{3}{16}$

Example 5: Extreme values · 5 marks · Long answerHOTS / competency

Find the greatest and least values of $3 sin x + 4 cos x$ and the value of $tan x$ at which the greatest value occurs.

Try it first, then show the solutionSolution

Write $3 sin x + 4 cos x = R sin (x + α) = R sin x cos α + R cos x sin α$ . (1 mark)
Compare: $R cos α = 3$ , $R sin α = 4$ , so $R = 9 + 16 = 5$ , $tan α = \frac{4}{3}$ . (1 mark)
Since $- 1 \leq sin (x + α) \leq 1$ , the expression lies between $- 5$ and $5$ . (1 mark)
Greatest value $5$ when $x + α = \frac{π}{2}$ , so $x = \frac{π}{2} - α$ and $sin x = cos α = \frac{3}{5}$ , $cos x = sin α = \frac{4}{5}$ , $tan x = \frac{3}{4}$ . (1 mark)
Check: $3 \cdot \frac{3}{5} + 4 \cdot \frac{4}{5} = \frac{25}{5} = 5$ . Least value $- 5$ . (1 mark)

Answer: Greatest 5, least $- 5$ ; greatest when $tan x = \frac{3}{4}$

Where students lose marks in Sum, Product and Multiple-Angle Identities with Proofs

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

Mistake: Wrong sign in the cosine of a sum
The sign between the products is opposite to the angle sign.
Wrong working
$cos (A + B) = cos A cos B + sin A sin B$ .
Right working
$cos (A + B) = cos A cos B - sin A sin B$ .
Do this: Remember "cos changes the sign".
Usually costs: The answer.
Mistake: Starting with the simpler side
There is nothing to expand, so the proof stalls.
Wrong working
Beginning with $4 cos x cos 2 x sin 4 x$ and expanding it.
Right working
Start from the sum of four sines and group in pairs.
Do this: Begin with the longer side.
Usually costs: Method marks.
Mistake: Pairing the wrong terms for sum-to-product
The pairs must give the same half-sum.
Wrong working
$sin x + sin 3 x$ and $sin 5 x + sin 7 x$ paired.
Right working
Pair $sin 7 x + sin x$ and $sin 5 x + sin 3 x$ (both half-sum $4 x$ ).
Do this: Check that pair sums match.
Usually costs: The proof.
Mistake: Forgetting the half angles
The sum-to-product formulas halve the sum and difference.
Wrong working
$sin A + sin B = 2 sin (A + B) cos (A - B)$ .
Right working
$2 sin \frac{A + B}{2} cos \frac{A - B}{2}$ .
Do this: Write the halves explicitly.
Usually costs: The answer.

Exam tips

Write the formula you use at each step; the marks follow the formulas.
Begin proofs from the side with more terms and state the target.
For exact values write the angle as a sum or difference of $3 0^{\circ}$ , $4 5^{\circ}$ , $6 0^{\circ}$ and $9 0^{\circ}$ first.
Test an identity at one angle (for example $x = 3 0^{\circ}$ ) as a quick check.
Remember that questions on $2 x$ and $3 x$ identities (and their proofs) are part of the 2026-27 syllabus.

Practice questions on Sum, Product and Multiple-Angle Identities with Proofs

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
The value of $cos 1 5^{\circ}$ is:
(a) $\frac{6 + 2}{4}$
(b) $\frac{6 - 2}{4}$
(c) $\frac{3 + 1}{2}$
(d) $\frac{3 - 1}{2}$
Hint
$1 5^{\circ} = 4 5^{\circ} - 3 0^{\circ}$ .
Answer
(a). $cos 4 5^{\circ} cos 3 0^{\circ} + sin 4 5^{\circ} sin 3 0^{\circ} = \frac{6 + 2}{4}$ .
Q2 · 1 mark · MCQHarder board level
The value of $cos 3 6^{\circ} cos 7 2^{\circ}$ is:
(a) $\frac{1}{4}$
(b) $\frac{1}{2}$
(c) $\frac{1}{8}$
(d) $\frac{5}{4}$
Hint
Multiply and divide by $2 sin 3 6^{\circ}$ .
Answer
(a). $\frac{2 sin 3 6 ^{\circ} cos 3 6 ^{\circ} cos 7 2 ^{\circ}}{2 sin 3 6 ^{\circ}} = \frac{sin 7 2 ^{\circ} cos 7 2 ^{\circ}}{2 sin 3 6 ^{\circ}} = \frac{sin 14 4 ^{\circ}}{4 sin 3 6 ^{\circ}} = \frac{1}{4}$ .
Q3 · 3 marks · Short answerHarder board level
If $cos x = - \frac{3}{5}$ and $π < x < \frac{3 π}{2}$ , find $sin 2 x$ , $cos 2 x$ and $tan 2 x$ .
Hint
Find $sin x$ with its sign first.
Answer
$sin x = - \frac{4}{5}$ . $sin 2 x = 2 (- \frac{4}{5}) (- \frac{3}{5}) = \frac{24}{25}$ ; $cos 2 x = 2 \cdot \frac{9}{25} - 1 = - \frac{7}{25}$ ; $tan 2 x = \frac{24/25}{- 7/25} = - \frac{24}{7}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $cos 3 x = 4 cos^{3} x - 3 cos x$ .
Reason (R): It follows by writing $cos 3 x = cos (2 x + x)$ and expanding with the sum formula and the double-angle formulae.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Expand $cos (2 x + x)$ .
Answer
(a). $cos 2 x cos x - sin 2 x sin x = (2 cos^{2} x - 1) cos x - 2 sin^{2} x cos x = 4 cos^{3} x - 3 cos x$ .
Q5 · 4 marks · Short answerHOTS / competency
Prove that $cos 2 0^{\circ} cos 4 0^{\circ} cos 8 0^{\circ} = \frac{1}{8}$ .
Hint
Multiply and divide by $8 sin 2 0^{\circ}$ .
Answer
For $cos 2 0^{\circ} cos 4 0^{\circ} cos 8 0^{\circ}$ , multiply and divide by $8 sin 2 0^{\circ}$ : $8 sin 2 0^{\circ} cos 2 0^{\circ} = 4 sin 4 0^{\circ}$ ; $4 sin 4 0^{\circ} cos 4 0^{\circ} = 2 sin 8 0^{\circ}$ ; $2 sin 8 0^{\circ} cos 8 0^{\circ} = sin 16 0^{\circ} = sin 2 0^{\circ}$ . So the product $= \frac{sin 2 0 ^{\circ}}{8 sin 2 0 ^{\circ}} = \frac{1}{8}$ .

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.What is sin(A + B)?Show answer
sin A cos B + cos A sin B.
2.What is cos 3x?Show answer
4cos³x − 3cos x.
3.What is sin A + sin B?Show answer
2 sin((A + B)/2) cos((A − B)/2).
4.What is the maximum of 3 sin x + 4 cos x?Show answer
5.
5.What is cos 2x in terms of sin x?Show answer
1 − 2 sin²x.

Sum, Product and Multiple-Angle Identities with Proofs: frequently asked questions

How do I choose between the three forms of cos 2x?

Pick the form that leaves the required function: use 1 − 2sin²x for sine, 2cos²x − 1 for cosine, and cos²x − sin²x for a difference of squares.

Why does the sum-to-product formula use half angles?

Because the average of A and B and half of their difference appear when you add the expansions of sin(P + Q) and sin(P − Q) with P = (A + B)/2 and Q = (A − B)/2.

Is the proof of the triple-angle formula needed?

Yes, deriving sin 3x and cos 3x from the sum formula is a standard board-level question.

How can I check an identity quickly?

Put in a convenient angle such as 30° or 45° and compare the two sides numerically.