MathsICSE Class 10 · Trigonometric Identities

Using Algebraic Identities in Trigonometry: Cubes, Squares and Fourth Powers (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

Treat $\sin A$ and $\cos A$ like $a$ and $b$: the sum of cubes gives $\sin^3 A + \cos^3 A = (\sin A + \cos A)(1 - \sin A\cos A)$, and $(a + b)^2 = a^2 + 2ab + b^2$ gives $(\sin A + \cos A)^2 = 1 + 2\sin A\cos A$.

Replace a with sin A and b with cos A, then use a² + b² = 1 to turn each algebra identity into a trig identity.

Illustration drawn for QuestionBuddy.

Understand Algebraic Identities in Trigonometry

The syllabus calls these 'simple algebraic trigonometric expressions'. Both 2024 Q3(i) and 2025 Q4(iii) (4 marks each) needed $a^{3} + b^{3}$ , and the examiners report that many candidates failed to apply it.

The algebra you need.

$(a + b)^{2} = a^{2} + 2 ab + b^{2}$ , $(a - b)^{2} = a^{2} - 2 ab + b^{2}$
$a^{2} - b^{2} = (a - b) (a + b)$ , so $a^{4} - b^{4} = (a^{2} - b^{2}) (a^{2} + b^{2})$
$a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$ , $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$
$a^{4} + b^{4} = (a^{2} + b^{2})^{2} - 2 a^{2} b^{2}$

**With $a = sin A$ , $b = cos A$ ** (so $a^{2} + b^{2} = 1$ ):

$(sin A \pm cos A)^{2} = 1 \pm 2 sin A cos A$
$sin^{3} A + cos^{3} A = (sin A + cos A) (1 - sin A cos A)$
$sin^{3} A - cos^{3} A = (sin A - cos A) (1 + sin A cos A)$
$sin^{4} A + cos^{4} A = 1 - 2 sin^{2} A cos^{2} A$

**With $sec$ and $tan$ ** (where $sec^{2} A - tan^{2} A = 1$ ): $sec^{4} A - tan^{4} A = sec^{2} A + tan^{2} A$ , and $(sec A + tan A) (sec A - tan A) = 1$ .

Finding values from a sum or difference. If $sin A + cos A = p$ , squaring gives $sin A cos A = \frac{p ^{2} - 1}{2}$ . Then

$sin^{3} A + cos^{3} A = p (1 - \frac{p ^{2} - 1}{2})$ ,
$(sin A - cos A)^{2} = 1 - 2 sin A cos A = 2 - p^{2}$ .

This lets you find values without knowing the angle.

Spotting the pattern. Cubes in a numerator and a matching linear factor in the denominator (as in $\frac{s i n ^{3} A + c o s ^{3} A}{s i n A + c o s A}$ ) are a strong hint to factorise first and cancel.

Algebraic Identities in Trigonometry: formulas to remember

Sum of cubes: $sin^{3} A + cos^{3} A = (sin A + cos A) (1 - sin A cos A)$; From $a^{3} + b^{3}$ and $sin^{2} + cos^{2} = 1$ .
Difference of cubes: $sin^{3} A - cos^{3} A = (sin A - cos A) (1 + sin A cos A)$; From $a^{3} - b^{3}$ .
Square of a sum: $(sin A + cos A)^{2} = 1 + 2 sin A cos A$; Gives $sin A cos A$ from $sin A + cos A$ .
Fourth powers: $sin^{4} A + cos^{4} A = 1 - 2 sin^{2} A cos^{2} A$; $(a^{2} + b^{2})^{2} - 2 a^{2} b^{2}$ .

How to solve Algebraic Identities in Trigonometry questions

Spot the algebraic pattern, factorise or expand, then replace $sin^{2} + cos^{2}$ (or $sec^{2} - tan^{2}$ ) by 1.

1Rewrite the expression with $a = sin A$ , $b = cos A$ (or $sec A$ , $tan A$ ) in your head.
2Recognise the pattern: square of a sum, difference of squares, sum or difference of cubes, fourth powers.
3Apply the algebraic identity.
4Replace $a^{2} + b^{2}$ by 1 (or $a^{2} - b^{2}$ by 1 for $sec$ , $tan$ ).
5Cancel common factors and simplify to the required side.

Algebraic Identities in Trigonometry 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.

Product from a sum

Suppose a robot arm 1 m long is raised to an angle $A$ . Its horizontal reach is $cos A$ m and its height is $sin A$ m, and the sensors show that reach + height = 1.3 m.

Find reach × height and the difference between them, without finding $A$ .

Try it first, then show the solutionSolution

$(sin A + cos A)^{2} = 1 + 2 sin A cos A$ : $1.69 = 1 + 2 sin A cos A$ , so $sin A cos A = 0.345$ .
$(sin A - cos A)^{2} = 1 - 0.69 = 0.31$ , so the difference is $0.31 \approx 0.56$ m.

Answer: Product 0.345 m²; difference about 0.56 m

Memory hook: Square the sum to get the product.

Fourth powers from the product

Imagine a calculation for a tilted solar panel that needs $sin^{4} A + cos^{4} A$ , and you know only that $sin A cos A = 0.4$ .

Find the value.

Try it first, then show the solutionSolution

$sin^{4} A + cos^{4} A = (sin^{2} A + cos^{2} A)^{2} - 2 sin^{2} A cos^{2} A$ .
$= 1 - 2 \times 0.16 = 0.68$ .

Answer: 0.68

Memory hook: Fourth powers = 1 − 2(product)².

Tricks to remember Algebraic Identities in Trigonometry

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

a and b in disguise
Write $a = sin A$ , $b = cos A$ ; then $a^{2} + b^{2} = 1$ makes every algebraic identity shorter.
Use it when: Spotting the pattern.
Cubes: same sign, opposite sign, always plus
$a^{3} \pm b^{3} = (a \pm b) (a^{2} \mp ab + b^{2})$ : the first bracket copies the sign, the middle term takes the opposite sign.
Use it when: Factorising sums and differences of cubes.
Square the sum
Given $sin A + cos A$ , square it at once to get $sin A cos A$ .
Use it when: Finding values without the angle.

Algebraic Identities in Trigonometry: solved examples

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

Example 1: Two squares added · 1 mark · MCQBoard level

$(sin A + cos A)^{2} + (sin A - cos A)^{2}$ equals:
(a) 0
(b) 1
(c) 2
(d) 4

Try it first, then show the solutionSolution

$(1 + 2 sin A cos A) + (1 - 2 sin A cos A) = 2$ .

Answer: (c)

Example 2: Fourth powers · 2 marks · Short answerBoard level

Prove that $sin^{4} A + cos^{4} A = 1 - 2 sin^{2} A cos^{2} A$ .

Try it first, then show the solutionSolution

LHS $= (sin^{2} A)^{2} + (cos^{2} A)^{2} = (sin^{2} A + cos^{2} A)^{2} - 2 sin^{2} A cos^{2} A$ .
$= 1 - 2 sin^{2} A cos^{2} A =$ RHS.

Answer: Proved

Example 3: Difference of cubes · 3 marks · Short answerHarder board level

Factorise $sin^{3} A - cos^{3} A$ . Hence prove that $\frac{s i n ^{3} A - c o s ^{3} A}{s i n A - c o s A} = 1 + sin A cos A$ (for $A \neq = 4 5^{\circ}$ ).

Try it first, then show the solutionSolution

$sin^{3} A - cos^{3} A = (sin A - cos A) (sin^{2} A + sin A cos A + cos^{2} A) = (sin A - cos A) (1 + sin A cos A)$ .
Divide by $sin A - cos A$ (non-zero as $A \neq = 4 5^{\circ}$ ): LHS $= 1 + sin A cos A =$ RHS.

Answer: $(sin A - cos A) (1 + sin A cos A)$ ; proved

Example 4: Sec and tan to the fourth · 3 marks · Short answerHarder board level

Prove that $sec^{4} A - sec^{2} A = tan^{4} A + tan^{2} A$ .

Try it first, then show the solutionSolution

LHS $= sec^{2} A (sec^{2} A - 1) = sec^{2} A tan^{2} A$ .
RHS $= tan^{2} A (tan^{2} A + 1) = tan^{2} A sec^{2} A$ .
Both equal $sec^{2} A tan^{2} A$ , so LHS = RHS.

Answer: Proved

Example 5: From one relation to another · 3 marks · Short answerHOTS / competency

If $sin A + cos A = 2 cos A$ , prove that $cos A - sin A = 2 sin A$ .

Try it first, then show the solutionSolution

From the given relation, $sin A = (2 - 1) cos A$ .
So $cos A = \frac{s i n A}{2 - 1} = \frac{( 2 + 1 ) s i n A}{( 2 - 1 ) ( 2 + 1 )} = (2 + 1) sin A$ (difference of squares in the denominator).
Hence $cos A - sin A = 2 sin A$ .

Answer: Proved

Example 6: Case study: trigonometry as algebra · 4 marks · Case-basedHOTS / competency

Let $x = sin A$ and $y = cos A$ for an acute angle $A$ with $sin A > cos A$ , and suppose $x + y = 1.2$ .
(a) Show that $x^{4} + 2 x^{2} y^{2} + y^{4} = 1$ for every $A$ .
(b) Find $x y$ and $x^{3} + y^{3}$ .
(c) Find $x - y$ to 4 decimal places.

Try it first, then show the solutionSolution

(a) $x^{4} + 2 x^{2} y^{2} + y^{4} = (x^{2} + y^{2})^{2} = 1^{2} = 1$ .
(b) $(x + y)^{2} = 1 + 2 x y$ : $1.44 = 1 + 2 x y$ , so $x y = 0.22$ . $x^{3} + y^{3} = (x + y) (1 - x y) = 1.2 \times 0.78 = 0.936$ .
(c) $(x - y)^{2} = 1 - 2 x y = 0.56$ , and $x > y$ , so $x - y = 0.56 \approx 0.7483$ .

Answer: (b) $x y = 0.22$ , $x^{3} + y^{3} = 0.936$ (c) about 0.7483

Where students lose marks in Algebraic Identities in Trigonometry

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

Mistake: Not factorising a sum of cubes
Expanding or cross-multiplying makes the working long and error-prone.
Wrong working
Taking the LCM in $\frac{s i n ^{3} θ + c o s ^{3} θ}{s i n θ + c o s θ} + sin θ cos θ$ and expanding.
Right working
$sin^{3} θ + cos^{3} θ = (sin θ + cos θ) (1 - sin θ cos θ)$ , cancel, add: 1.
Do this: Look for cubes before anything else.
Usually costs: Reported in 2024 Q3(i) and 2025 Q4(iii).
Mistake: Middle term sign in the cube identity
$a^{3} + b^{3}$ has $- ab$ in the second bracket; $a^{3} - b^{3}$ has $+ ab$ .
Wrong working
$sin^{3} A + cos^{3} A = (sin A + cos A) (1 + sin A cos A)$
Right working
$(sin A + cos A) (1 - sin A cos A)$
Do this: Check by expanding one term mentally.
Usually costs: The whole proof after that line.
Mistake: Forgetting the cross term when squaring
$(sin A + cos A)^{2}$ is not $sin^{2} A + cos^{2} A$ .
Wrong working
$(sin A + cos A)^{2} = 1$
Right working
$(sin A + cos A)^{2} = 1 + 2 sin A cos A$
Do this: Write $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ before substituting.
Usually costs: Value questions built on $\sin A + \cos A$.

Exam tips

2024 Q3(i) (4 marks): factorise $sin^{3} θ + cos^{3} θ = (sin θ + cos θ) (1 - sin θ cos θ)$ , then the identity is $1 - sin θ cos θ + sin θ cos θ = 1$ .
2025 Q4(iii) (4 marks): $sin^{3} A + cos^{3} A$ appeared in the denominator and the numerator reduced to $\frac{s i n ^{3} A + c o s ^{3} A}{s i n A c o s A}$ ; cancelling gave $sec A cosec A$ .
The examiners suggest using previously learnt algebra to reach new results: revise the cube and square identities before the exam.
If the question says 'hence', use the factorisation you have just found.

Practice questions on Algebraic Identities in Trigonometry

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
$(1 - sin A) (1 + sin A)$ equals:
(a) $sin^{2} A$
(b) $cos^{2} A$
(c) $tan^{2} A$
(d) 1
Hint
Difference of squares.
Answer
(b)
Q2 · 2 marks · Short answerBoard level
Prove that $(1 + tan^{2} A) (1 - sin A) (1 + sin A) = 1$ .
Hint
Pair the last two brackets first.
Answer
LHS $= sec^{2} A (1 - sin^{2} A) = sec^{2} A cos^{2} A = 1$ .
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $sin^{3} A + cos^{3} A = (sin A + cos A) (1 - sin A cos A)$ .
Reason (R): $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Put $a = sin A$ , $b = cos A$ .
Answer
(c)
Q4 · 3 marks · Short answerHarder board level
Prove that $tan^{2} A - sin^{2} A = tan^{2} A sin^{2} A$ .
Hint
Take $sin^{2} A$ out as a common factor.
Answer
LHS $= \frac{s i n ^{2} A}{c o s ^{2} A} - sin^{2} A = sin^{2} A (\frac{1 - c o s ^{2} A}{c o s ^{2} A}) = sin^{2} A \times \frac{s i n ^{2} A}{c o s ^{2} A} = sin^{2} A tan^{2} A$ .
Q5 · 3 marks · Short answerHarder board level
Prove that $\frac{1 + c o s A}{1 - c o s A} - \frac{1 - c o s A}{1 + c o s A} = 4 cot A cosec A$ .
Hint
$(a + b)^{2} - (a - b)^{2} = 4 ab$ .
Answer
LHS $= \frac{( 1 + c o s A ) ^{2} - ( 1 - c o s A ) ^{2}}{1 - c o s ^{2} A} = \frac{4 c o s A}{s i n ^{2} A} = 4 \cdot \frac{c o s A}{s i n A} \cdot \frac{1}{s i n A} = 4 cot A cosec A$ .
Q6 · 3 marks · Short answerHarder board level
If $sin A - cos A = 0.2$ for an acute angle $A$ , find $sin A cos A$ , $sin^{3} A - cos^{3} A$ and $sin A + cos A$ .
Hint
Square the given relation.
Answer
$0.04 = 1 - 2 sin A cos A$ , so $sin A cos A = 0.48$ . $sin^{3} A - cos^{3} A = 0.2 \times 1.48 = 0.296$ . $(sin A + cos A)^{2} = 1.96$ , so $sin A + cos A = 1.4$ .
Q7 · 4 marks · Long answerHOTS / competency
Prove that $2 (sin^{6} A + cos^{6} A) - 3 (sin^{4} A + cos^{4} A) + 1 = 0$ .
Hint
Write $sin^{6} A + cos^{6} A$ as a sum of cubes of $sin^{2} A$ and $cos^{2} A$ .
Answer
Let $p = sin^{2} A cos^{2} A$ . $sin^{6} A + cos^{6} A = (sin^{2} A + cos^{2} A)^{3} - 3 p (sin^{2} A + cos^{2} A) = 1 - 3 p$ and $sin^{4} A + cos^{4} A = 1 - 2 p$ . LHS $= 2 - 6 p - 3 + 6 p + 1 = 0$ .

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.Factorise $sin^{3} A + cos^{3} A$ .Show answer
$(sin A + cos A) (1 - sin A cos A)$
2. $(sin A - cos A)^{2}$ ?Show answer
$1 - 2 sin A cos A$
3. $sin^{4} A + cos^{4} A$ ?Show answer
$1 - 2 sin^{2} A cos^{2} A$
4. $sin A + cos A = 1.4$ . $sin A cos A$ ?Show answer
0.48
5. $sec^{4} A - tan^{4} A$ ?Show answer
$sec^{2} A + tan^{2} A$

Algebraic Identities in Trigonometry: frequently asked questions

Why are algebraic identities used in trigonometry?

Expressions such as sin³A + cos³A or sin⁴A + cos⁴A are just algebra in sin A and cos A. Factorising them with a³ + b³ or a² − b² and then using sin²A + cos²A = 1 shortens the proof.

How do I factorise sin³A + cos³A?

Use a³ + b³ = (a + b)(a² − ab + b²): it becomes (sin A + cos A)(sin²A − sin A cos A + cos²A) = (sin A + cos A)(1 − sin A cos A).

How can I find sin A cos A if sin A + cos A is known?

Square it: (sin A + cos A)² = 1 + 2 sin A cos A, so sin A cos A = (p² − 1)/2 where p = sin A + cos A.