Trigonometric Ratios And Identities

Q: What are the compound angle formulas?

The main ones are sin(A ± B) = sin A cos B ± cos A sin B and cos(A ± B) = cos A cos B ∓ sin A sin B. From these, tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B) follows by division.

Q: What is the formula for sin 2A?

sin 2A = 2 sin A cos A. It follows from sin(A+A) using the compound-angle formula. Similarly cos 2A = cos² A - sin² A = 2cos² A - 1 = 1 - 2sin² A.

Q: How do you convert sum to product in trigonometry?

Sum-to-product formulas: sin C + sin D = 2 sin((C+D)/2) cos((C-D)/2); cos C + cos D = 2 cos((C+D)/2) cos((C-D)/2). They are derived by adding compound-angle formulas.

Q: How to find the maximum and minimum value of a sin x + b cos x?

Rewrite as R sin(x + φ) where R = √(a² + b²) and tan φ = b/a. Since sin ranges over [-1, 1], the expression ranges over [-√(a² + b²), √(a² + b²)].

Q: What is the value of sin 15° and cos 15°?

sin 15° = (√3 - 1)/(2√2) and cos 15° = (√3 + 1)/(2√2). These come from sin(45° - 30°) and cos(45° - 30°).

Q: What is the difference between sin 2A and 2 sin A?

sin 2A = 2 sin A cos A, not 2 sin A. The factor of cos A is crucial. For example sin 60° = √3/2, whereas 2 sin 30° = 1; but 2 sin 30° cos 30° = 2 · 1/2 · √3/2 = √3/2.

Q: How do I derive tan 3A?

Use tan 3A = tan(2A + A) and the compound formula: tan 3A = (tan 2A + tan A)/(1 - tan 2A tan A). Substituting tan 2A = 2 tan A/(1 - tan² A) and simplifying gives tan 3A = (3 tan A - tan³ A)/(1 - 3 tan² A).

MathsTrigonometryFor JEE aspirants

Trigonometric identities are equations connecting trigonometric functions that hold for every value of the variable. This page covers compound-angle formulas $(sin (A \pm B), cos (A \pm B), tan (A \pm B))$ , transformation formulas (sum-to-product and product-to-sum), multiple- and sub-multiple-angle formulas ( $sin 2 A$ , $cos 2 A$ , $sin 3 A$ , $cos 3 A$ , $tan A /2$ ), conditional identities when $A + B + C = π$ , sums of trigonometric ratios in arithmetic progression, and the maximum-minimum range of $a sin x + b cos x$ - the identity toolkit that JEE aspirants need for every trig problem beyond the definitions.

Key Formulas - Quick Reference

$sin (A \pm B) = sin A cos B \pm cos A sin B$ ; $cos (A \pm B) = cos A cos B \mp sin A sin B$
$tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B}$
$sin 2 A = 2 sin A cos A = \frac{2 tan A}{1 + tan ^{2} A}$ ; $cos 2 A = 1 - 2 sin^{2} A = 2 cos^{2} A - 1 = \frac{1 - tan ^{2} A}{1 + tan ^{2} A}$
$sin 3 A = 3 sin A - 4 sin^{3} A$ ; $cos 3 A = 4 cos^{3} A - 3 cos A$ ; $tan 3 A = \frac{3 tan A - tan ^{3} A}{1 - 3 tan ^{2} A}$
$sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2}$ ; $cos C + cos D = 2 cos \frac{C + D}{2} cos \frac{C - D}{2}$
Range: $- a^{2} + b^{2} \leq a sin x + b cos x \leq a^{2} + b^{2}$

1. Compound Angle Formulas

The two fundamental compound-angle identities express $sin (A \pm B)$ and $cos (A \pm B)$ in terms of ratios of $A$ and $B$ :

sin (A + B) = sin A cos B + cos A sin B

sin (A - B) = sin A cos B - cos A sin B

cos (A + B) = cos A cos B - sin A sin B

cos (A - B) = cos A cos B + sin A sin B

A quick geometric picture of $sin (A + B)$ :

Figure: Geometric proof of

sin (A + B) = sin A cos B + cos A sin B

1.1 Tangent and Cotangent of Sum/Difference

tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B} cot (A \pm B) = \frac{cot A cot B \mp 1}{cot B \pm cot A}

1.2 Squared Sum-Difference Identities

sin^{2} A - sin^{2} B = sin (A + B) sin (A - B) = cos^{2} B - cos^{2} A

cos^{2} A - sin^{2} B = cos (A + B) cos (A - B) = cos^{2} B - sin^{2} A

1.3 Three-Angle Sum

sin (A + B + C) = sin A cos B cos C + sin B cos A cos C + sin C cos A cos B - sin A sin B sin C

cos (A + B + C) = cos A cos B cos C - cos A sin B sin C - sin A cos B sin C - sin A sin B cos C

tan (A + B + C) = \frac{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}

1.4 $n$ -Angle Sum for Tangent

JEE Advanced

tan (θ_{1} + θ_{2} + \dots + θ_{n}) = \frac{S _{1} - S _{3} + S _{5} - \dots}{1 - S _{2} + S _{4} - \dots}

where

S_{k}

= sum of products of

tan θ_{i}

taken

k

at a time.

Solved Example 1

Prove

sin (45° + A) cos (45° - B) + cos (45° + A) sin (45° - B) = cos (A - B)

Solution:

The LHS matches $sin X cos Y + cos X sin Y = sin (X + Y)$ with $X = 45° + A$ , $Y = 45° - B$ :

LHS = sin ((45° + A) + (45° - B)) = sin (90° + (A - B)) = cos (A - B) .

Solved Example 2

Prove

tan (\frac{π}{4} + θ) \cdot tan (\frac{3 π}{4} + θ) = - 1

Solution:

$tan (\frac{π}{4} + θ) = \frac{1 + tan θ}{1 - tan θ}$ .

$tan (\frac{3 π}{4} + θ) = tan (π - (\frac{π}{4} - θ)) = - tan (\frac{π}{4} - θ) = - \frac{1 - tan θ}{1 + tan θ}$ .

Their product simplifies to $- 1$ .

2. Transformation Formulas

2.1 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)

2.2 Sum $\to$ 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{D - C}{2}

Solved Example 3

Prove

sin 5 A + sin 3 A = 2 sin 4 A cos A

Solution:

Using $sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2}$ with $C = 5 A, D = 3 A$ :

LHS = 2 sin \frac{5 A + 3 A}{2} cos \frac{5 A - 3 A}{2} = 2 sin 4 A cos A .

Solved Example 4

Prove

\frac{sin 8 A cos A - sin 6 A cos 3 A}{cos 2 A cos A - sin 3 A sin 4 A} = tan 2 A

Solution:

Multiply numerator and denominator by $2$ and convert each product to sums:

Num = \frac{1}{2} [(s in 9 A + sin 7 A) - (sin 9 A + sin 3 A)] = \frac{1}{2} (sin 7 A - sin 3 A)

Not quite - better to group differently. Numerator $\times 2 = (sin 9 A + sin 7 A) - (sin 9 A + sin 3 A) = sin 7 A - sin 3 A = 2 cos 5 A sin 2 A$ .

Denominator $\times 2 = (cos 3 A + cos A) - (cos A - cos 7 A) = cos 3 A + cos 7 A = 2 cos 5 A cos 2 A$ .

Ratio $= \frac{2 cos 5 A sin 2 A}{2 cos 5 A cos 2 A} = tan 2 A$ .

3. Multiple-Angle Formulas

3.1 Double-Angle Formulas

sin 2 A = 2 sin A cos A = \frac{2 tan A}{1 + tan ^{2} A}

cos 2 A = cos^{2} A - sin^{2} A = 2 cos^{2} A - 1 = 1 - 2 sin^{2} A = \frac{1 - tan ^{2} A}{1 + tan ^{2} A}

tan 2 A = \frac{2 tan A}{1 - tan ^{2} A}

Rearrangements often used to lower powers:

2 cos^{2} A = 1 + cos 2 A, 2 sin^{2} A = 1 - cos 2 A

3.2 Triple-Angle Formulas

sin 3 A = 3 sin A - 4 sin^{3} A

cos 3 A = 4 cos^{3} A - 3 cos A

tan 3 A = \frac{3 tan A - tan ^{3} A}{1 - 3 tan ^{2} A}

Solved Example 5

Prove

tan A + cot A = 2 csc 2 A

Solution:

$tan A + cot A = \frac{sin A}{cos A} + \frac{cos A}{sin A} = \frac{sin ^{2} A + cos ^{2} A}{sin A cos A} = \frac{1}{sin A cos A} = \frac{2}{2 sin A cos A} = \frac{2}{sin 2 A} = 2 csc 2 A$ .

Solved Example 6

Prove

sin 20° sin 40° sin 60° sin 80° = \frac{3}{16}

Solution:

Group $sin 20° sin 40° sin 80° = sin 20° sin (60° - 20°) sin (60° + 20°) = \frac{1}{4} sin 3 \cdot 20° = \frac{1}{4} sin 60° = \frac{3}{8}$ .

(Using $sin θ sin (60° - θ) sin (60° + θ) = \frac{1}{4} sin 3 θ$ .)

Multiplying by $sin 60° = \frac{3}{2}$ gives $\frac{3}{8} \cdot \frac{3}{2} = \frac{3}{16}$ .

4. Sub-Multiple Angle Formulas ( $A /2$ )

Setting $A \to A /2$ in the double-angle formulas gives half-angle relations:

sin A = 2 sin \frac{A}{2} cos \frac{A}{2}

cos A = 1 - 2 sin^{2} \frac{A}{2} = 2 cos^{2} \frac{A}{2} - 1

tan A = \frac{2 tan ( A /2 )}{1 - tan ^{2} ( A /2 )}

Useful "half-angle in surd form" identities:

sin \frac{A}{2} = \pm \frac{1 - cos A}{2}, cos \frac{A}{2} = \pm \frac{1 + cos A}{2}, tan \frac{A}{2} = \pm \frac{1 - cos A}{1 + cos A}

Signs are chosen by the quadrant of $A /2$ .

5. Values at Special Angles

Angle	Sine	Cosine	Tangent
$15°$ or $π /12$	$\frac{3 - 1}{2 2}$	$\frac{3 + 1}{2 2}$	$2 - 3$
$18°$ or $π /10$	$\frac{5 - 1}{4}$	$\frac{10 + 2 5}{4}$	-
$36°$ or $π /5$	$\frac{10 - 2 5}{4}$	$\frac{5 + 1}{4}$	-
$75°$ or $5 π /12$	$\frac{3 + 1}{2 2}$	$\frac{3 - 1}{2 2}$	$2 + 3$

Also useful: $sin (nπ) = 0$ , $cos (nπ) = (- 1)^{n}$ , $tan (nπ) = 0$ for every integer $n$ .

6. Conditional Identities (when $A + B + C = π$ )

If $A, B, C$ are the angles of a triangle then several identities hold:

Triangle Identities (

A + B + C = π

)

$sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C$
$cos 2 A + cos 2 B + cos 2 C = - 1 - 4 cos A cos B cos C$
$sin A + sin B + sin C = 4 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}$
$cos A + cos B + cos C = 1 + 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}$
$tan A + tan B + tan C = tan A tan B tan C$
$cot A cot B + cot B cot C + cot C cot A = 1$
$tan \frac{A}{2} tan \frac{B}{2} + tan \frac{B}{2} tan \frac{C}{2} + tan \frac{C}{2} tan \frac{A}{2} = 1$
$cot \frac{A}{2} + cot \frac{B}{2} + cot \frac{C}{2} = cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}$

If instead $A + B + C = π /2$ , then $tan A tan B + tan B tan C + tan C tan A = 1$ .

Solved Example 7

A + B + C = 180°

, prove

sin^{2} A + sin^{2} B + sin^{2} C = 2 + 2 cos A cos B cos C

Solution:

Let $S = sin^{2} A + sin^{2} B + sin^{2} C$ .

$2 S = 2 sin^{2} A + (1 - cos 2 B) + (1 - cos 2 C) = 2 sin^{2} A + 2 - (cos 2 B + cos 2 C)$ .

$cos 2 B + cos 2 C = 2 cos (B + C) cos (B - C) = - 2 cos A cos (B - C)$ (since $B + C = π - A$ ).

So $2 S = 2 sin^{2} A + 2 + 2 cos A cos (B - C) = (2 - 2 cos^{2} A) + 2 + 2 cos A cos (B - C)$ .

$= 4 - 2 cos A [cos A - cos (B - C)] = 4 - 2 cos A [- cos (B + C) - cos (B - C)] \cdot (- 1)$ … simplify to $= 4 + 4 cos A cos B cos C$ . Divide by $2$ : $S = 2 + 2 cos A cos B cos C$ .

7. Series - Sines and Cosines in Arithmetic Progression

For angles in AP with first term $α$ and common difference $β$ (with $β \neq = 2 mπ$ ):

sin α + sin (α + β) + sin (α + 2 β) + \dots + sin (α + (n - 1) β) = \frac{sin \frac{n β}{2}}{sin \frac{β}{2}} \cdot sin (α + \frac{( n - 1 ) β}{2})

cos α + cos (α + β) + cos (α + 2 β) + \dots + cos (α + (n - 1) β) = \frac{sin \frac{n β}{2}}{sin \frac{β}{2}} \cdot cos (α + \frac{( n - 1 ) β}{2})

7.1 Product of Cosines with Doubling Angles

cos θ \cdot cos 2 θ \cdot cos 2^{2} θ \dots cos 2^{n - 1} θ = \frac{sin 2 ^{n} θ}{2 ^{n} sin θ}

Solved Example 8

Find the value of

cos \frac{π}{7} + cos \frac{2 π}{7} + cos \frac{3 π}{7} + cos \frac{4 π}{7} + cos \frac{5 π}{7} + cos \frac{6 π}{7}

Solution:

An AP with $α = π /7$ , $β = π /7$ , $n = 6$ . Using the cosine-series formula:

Sum $= \frac{sin ( 6 π /14 )}{sin ( π /14 )} \cdot cos (π /7 + 5 π /14) = \frac{sin ( 3 π /7 )}{sin ( π /14 )} \cdot cos (π /2) = 0$ (since $cos (π /2) = 0$ ).

8. Range of $a sin x + b cos x$

Write $E = a sin x + b cos x$ and set $R = a^{2} + b^{2}$ . Then

E = R (\frac{a}{R} sin x + \frac{b}{R} cos x) = R (cos φ sin x + sin φ cos x) = R sin (x + φ)

where $tan φ = b / a$ . Since $sin (x + φ) \in [- 1, 1]$ :

- a^{2} + b^{2} \leq a sin x + b cos x \leq a^{2} + b^{2}

Maximum

a^{2} + b^{2}

attained when

sin (x + φ) = 1

; minimum

- a^{2} + b^{2}

when

sin (x + φ) = - 1

Figure:

a sin x + b cos x = R sin (x + φ)

oscillates between

\pm a^{2} + b^{2}

Solved Example 9

Find the maximum and minimum values of

3 sin x + 4 cos x

Solution:

$R = 3^{2} + 4^{2} = 5$ . So $- 5 \leq 3 sin x + 4 cos x \leq 5$ ; max $= 5$ , min $= - 5$ .

Solved Example 10

Find the maximum value of

1 + 2 sin x + 3 cos^{2} x

Solution:

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

Let $t = sin x$ so $t \in [- 1, 1]$ and $f (t) = - 3 t^{2} + 2 t + 4 = - 3 (t - \frac{1}{3})^{2} + \frac{13}{3}$ .

Max at $t = 1/3$ : $f_{m a x} = \frac{13}{3}$ . Min at $t = - 1$ : $f_{m i n} = - 3 - 2 + 4 = - 1$ .

Common Mistakes to Avoid

Watch out

Sign error in $cos (A + B)$ . The formula is $cos A cos B - sin A sin B$ (minus, not plus). A common slip is to write plus by analogy with $sin (A + B)$ .
Applying $tan (A + B)$ when $A + B = π /2$ . Then $tan A tan B = 1$ making the denominator zero. The formula fails; use another approach.
Range of $a sin x + b cos x + c$ . The constant $c$ shifts the range: it is $[c - a^{2} + b^{2}, c + a^{2} + b^{2}]$ . Don't forget the $c$ .
Assuming half-angle formulas keep the positive sign. The $\pm$ signs depend on the quadrant of $A /2$ , which need not be the quadrant of $A$ .
Confusing $sin C - sin D$ with $sin C + sin D$ . $sin C - sin D = 2 cos \frac{C + D}{2} sin \frac{C - D}{2}$ - the cosine and sine swap places compared to the sum formula.
Forgetting conditional identities have a condition. Identities like $tan A + tan B + tan C = tan A tan B tan C$ hold only when $A + B + C = π$ (or an integer multiple), not in general.

Frequently Asked Questions

What are the compound angle formulas?

The main ones are $sin (A \pm B) = sin A cos B \pm cos A sin B$ and $cos (A \pm B) = cos A cos B \mp sin A sin B$ . From these, $tan (A \pm B) = (tan A \pm tan B) / (1 \mp tan A tan B)$ follows by division.

What is the formula for sin 2A?

$sin 2 A = 2 sin A cos A$ . It follows from $sin (A + A)$ using the compound-angle formula. Similarly $cos 2 A = cos^{2} A - sin^{2} A = 2 cos^{2} A - 1 = 1 - 2 sin^{2} A$ .

How do you convert sum to product in trigonometry?

Sum-to-product formulas: $sin C + sin D = 2 sin (\frac{C + D}{2}) cos (\frac{C - D}{2})$ ; $cos C + cos D = 2 cos (\frac{C + D}{2}) cos (\frac{C - D}{2})$ . They are derived by adding compound-angle formulas.

What are conditional identities in trigonometry?

Identities that hold only when the angles satisfy a condition, most commonly $A + B + C = π$ (angles of a triangle). Example: if $A + B + C = π$ then $tan A + tan B + tan C = tan A tan B tan C$ .

How to find the maximum and minimum value of a sin x + b cos x?

Rewrite as $R sin (x + φ)$ where $R = a^{2} + b^{2}$ and $tan φ = b / a$ . Since $sin$ ranges over $[- 1, 1]$ , the expression ranges over $[- a^{2} + b^{2}, a^{2} + b^{2}]$ .

What is the value of sin 15° and cos 15°?

$sin 15° = \frac{3 - 1}{2 2}$ and $cos 15° = \frac{3 + 1}{2 2}$ . These come from $sin (45° - 30°)$ and $cos (45° - 30°)$ .

What is the difference between sin 2A and 2 sin A?

$sin 2 A = 2 sin A cos A$ , not $2 sin A$ . The factor of $cos A$ is crucial. For example $sin 60° = 3 /2$ , whereas $2 sin 30° = 1$ ; but $2 sin 30° cos 30° = 2 \cdot 1/2 \cdot 3 /2 = 3 /2$ .

How do I derive tan 3A?

Use $tan 3 A = tan (2 A + A)$ and the compound formula: $tan 3 A = \frac{tan 2 A + tan A}{1 - tan 2 A tan A}$ . Substituting $tan 2 A = 2 tan A / (1 - tan^{2} A)$ and simplifying gives $tan 3 A = \frac{3 tan A - tan ^{3} A}{1 - 3 tan ^{2} A}$ .

Previous year questions on Trigonometric Ratios And Identities

19 questions from past papers, each with a step-by-step solution.

Show all 19 questions

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1. Compound Angle Formulas

1.1 Tangent and Cotangent of Sum/Difference

1.2 Squared Sum-Difference Identities

1.3 Three-Angle Sum

1.4 n-Angle Sum for Tangent

2. Transformation Formulas

2.1 Product → Sum

2.2 Sum → Product

3. Multiple-Angle Formulas

3.1 Double-Angle Formulas

3.2 Triple-Angle Formulas

4. Sub-Multiple Angle Formulas (A/2)

5. Values at Special Angles

6. Conditional Identities (when A+B+C=π)

7. Series - Sines and Cosines in Arithmetic Progression

7.1 Product of Cosines with Doubling Angles

8. Range of asinx+bcosx

Common Mistakes to Avoid

Frequently Asked Questions

What are the compound angle formulas?

What is the formula for sin 2A?

How do you convert sum to product in trigonometry?

What are conditional identities in trigonometry?

How to find the maximum and minimum value of a sin x + b cos x?

What is the value of sin 15° and cos 15°?

What is the difference between sin 2A and 2 sin A?

How do I derive tan 3A?

Previous year questions on Trigonometric Ratios And Identities

Ready to master Trigonometry?

1.4 $n$ -Angle Sum for Tangent

2.1 Product $\to$ Sum

2.2 Sum $\to$ Product

4. Sub-Multiple Angle Formulas ( $A /2$ )

6. Conditional Identities (when $A + B + C = π$ )

8. Range of $a sin x + b cos x$