Module 1 · NDA Mathematics (Paper 1)

Trigonometry

Trigonometric ratios, identities, equations, inverse trig, height & distance.

Trig Ratios · Identities · Equations · Inverse Trig

🎯 Take Mock Test

1. Angles & Trigonometric Ratios

Trigonometry is the study of relationships between angles and sides of triangles. For NDA Paper 1, approximately 6 to 8 questions (15 to 20 marks) appear from this topic. Mastery of angle measure, standard ratios, and the sign convention is essential for scoring full marks in this section. Trigonometry is also a prerequisite for calculus and coordinate geometry topics in the NDA syllabus.

Degree & Radian Measure

Angles can be measured in two systems: the sexagesimal system (degrees, minutes, seconds) and the circular system (radians). The fundamental relationship that links the two systems is based on the fact that a full circle measures 360 degrees or 2π radians.

π radians = 180° → 1 radian = 180°/π ≈ 57.2958°

To convert from degrees to radians, multiply the degree measure by π/180. To convert from radians to degrees, multiply the radian measure by 180/π.

Common angle conversions that every NDA candidate must memorise: 0° = 0 rad, 30° = π/6 rad, 45° = π/4 rad, 60° = π/3 rad, 90° = π/2 rad, 180° = π rad, 270° = 3π/2 rad, 360° = 2π rad. Additionally, 15° = π/12 rad, 75° = 5π/12 rad, and 120° = 2π/3 rad are also useful.

Arc Length and Sector Area

In a circle of radius r, if an angle θ (measured in radians) subtends an arc at the centre, the length of the arc is given by L = rθ. The area of the corresponding sector is A = ½ r²θ. These formulas are valid only when θ is expressed in radians, not degrees.

For NDA, problems involving arc length or sector area typically give the angle in degrees first, requiring conversion to radians before applying the formulas. The perimeter of a sector is 2r + rθ.

NDA Shortcut

Memorise the radian equivalents of 0°, 30°, 45°, 60°, 90°, 180°, 270°, 360°. NDA often asks direct conversion questions (e.g., "Convert 150° to radians" or "Convert 3π/4 to degrees"). The pattern 0, π/6, π/4, π/3, π/2, π, 3π/2, 2π must be on your fingertips for instant recall.

Example 1 — Degree-Radian Conversion

(a) Convert 150° to radians. (b) Convert 5π/4 radians to degrees. (c) What is the radian measure of 240°?

Solution: (a) 150° × (π/180) = 150π/180 = 5π/6 rad. (b) 5π/4 × (180/π) = (5 × 180)/4 = 900/4 = 225°. (c) 240° × (π/180) = 4π/3 rad.

Example 2 — Arc Length

Find the length of an arc of a circle of radius 14 cm that subtends an angle of 60° at the centre.

Solution: First convert 60° to radians: 60° = π/3 rad. Arc length L = rθ = 14 × π/3 = 14π/3 cm ≈ 14.67 cm.

Trigonometric Ratios and Identities

The six fundamental trigonometric ratios are defined using the unit circle (a circle of radius 1 centred at the origin) or a right-angled triangle. On the unit circle, the x-coordinate gives cos θ, the y-coordinate gives sin θ, and their ratio gives tan θ.

sin θ = opposite side / hypotenuse = y-coordinate on the unit circle
cos θ = adjacent side / hypotenuse = x-coordinate on the unit circle
tan θ = sin θ / cos θ = opposite side / adjacent side
cot θ = cos θ / sin θ = 1 / tan θ
sec θ = 1 / cos θ = hypotenuse / adjacent side
cosec θ = 1 / sin θ = hypotenuse / opposite side

Standard Angle Values

The following table of standard angle values must be memorised completely. NDA frequently tests these values directly as well as in compound-angle and multiple-angle problems. Pay special attention to the pattern: sin increases from 0 to 1 as cos decreases from 1 to 0.

Angle θ	0° (0)	30° (π/6)	45° (π/4)	60° (π/3)	90° (π/2)	180° (π)	270° (3π/2)	360° (2π)
sin θ	0	1/2	1/√2	√3/2	1	0	−1	0
cos θ	1	√3/2	1/√2	1/2	0	−1	0	1
tan θ	0	1/√3	1	√3	∞ (undefined)	0	∞ (undefined)	0

NDA Shortcut

A quick way to remember sin values for 0°, 30°, 45°, 60°, 90°: write √0/2 = 0, √1/2 = 1/2, √2/2 = 1/√2, √3/2, √4/2 = 1. Cos values are the same sequence in reverse. This pattern saves you from rote memorisation and helps in solving NDA problems faster.

Complementary Angle Relations

When two angles add up to 90° (π/2), they are called complementary. The trigonometric ratios of complementary angles follow a simple interchange pattern:

Complementary Pair	Relation
sin(90° − θ)	= cos θ
cos(90° − θ)	= sin θ
tan(90° − θ)	= cot θ
cot(90° − θ)	= tan θ
sec(90° − θ)	= cosec θ
cosec(90° − θ)	= sec θ

Fundamental Trigonometric Identities

The three Pythagorean identities form the foundation of all trigonometric simplifications:

sin²θ + cos²θ = 1 — Dividing by cos²θ gives 1 + tan²θ = sec²θ
1 + tan²θ = sec²θ — Valid for all θ where cos θ ≠ 0
1 + cot²θ = cosec²θ — Valid for all θ where sin θ ≠ 0

Reciprocal Identities: sin θ · cosec θ = 1, cos θ · sec θ = 1, tan θ · cot θ = 1.

Quotient Identities: tan θ = sin θ / cos θ, cot θ = cos θ / sin θ.

NDA Shortcut

The identity sin²θ + cos²θ = 1 is the single most used trigonometric identity in NDA. Whenever you see an expression like 1 − sin²θ or 1 − cos²θ, immediately replace it with cos²θ or sin²θ respectively. To derive the other two Pythagorean identities, simply divide sin² + cos² = 1 by cos² or sin².

Example 3 — NDA 2020

If sin θ = 3/5 and θ is acute, find the value of tan θ + sec θ.

Solution: Using sin²θ + cos²θ = 1, we get cos²θ = 1 − 9/25 = 16/25. Therefore cos θ = 4/5 (positive since θ is acute). Then tan θ = sin θ / cos θ = (3/5)/(4/5) = 3/4. Sec θ = 1 / cos θ = 5/4. The required sum = 3/4 + 5/4 = 8/4 = 2. This is a classic NDA problem that tests the fundamental identity and ratio definitions together.

Example 4 — Identity Proof

Prove that (1 + tan²θ)(1 − sin θ)(1 + sin θ) = 1.

Solution: (1 + tan²θ) = sec²θ. (1 − sin θ)(1 + sin θ) = 1 − sin²θ = cos²θ. Therefore LHS = sec²θ · cos²θ = (1/cos²θ) · cos²θ = 1 = RHS. Hence proved.

Sign Convention (ASTC Rule)

The sign of each trigonometric function depends on the quadrant in which the terminal side of the angle lies. The ASTC rule helps in remembering which functions are positive in each quadrant. The mnemonic stands for All, Sin, Tan, Cos — going counterclockwise from Quadrant I.

Quadrant I (0° to 90°): All six trigonometric functions are positive.
Quadrant II (90° to 180°): Only Sin and cosec are positive; cos, tan, cot, sec are negative.
Quadrant III (180° to 270°): Only Tan and cot are positive; sin, cos, sec, cosec are negative.
Quadrant IV (270° to 360°): Only Cos and sec are positive; sin, tan, cot, cosec are negative.

Trigonometric Functions of Allied Angles

Allied angles are angles of the form nπ/2 ± θ where n is an integer. The trigonometric ratios of such angles can be reduced to those of θ using the following rules. Note the pattern: if n is odd, the function co-functions (sin ↔ cos, tan ↔ cot, sec ↔ cosec); if n is even, the function stays the same. The sign is determined by the ASTC rule for the quadrant in which the angle lies.

Angle	sin	cos	tan
−θ	−sin θ	cos θ	−tan θ
90° − θ	cos θ	sin θ	cot θ
90° + θ	cos θ	−sin θ	−cot θ
180° − θ	sin θ	−cos θ	−tan θ
180° + θ	−sin θ	−cos θ	tan θ
270° − θ	−cos θ	−sin θ	cot θ
270° + θ	−cos θ	sin θ	−cot θ
360° − θ	−sin θ	cos θ	−tan θ
360° + θ	sin θ	cos θ	tan θ

NDA Shortcut

For allied angles of the form (90°n ± θ): Step 1 — If n is odd, change the function (sin ↔ cos, tan ↔ cot, sec ↔ cosec). If n is even, keep the same function. Step 2 — Determine the quadrant of the original angle and assign the correct sign using ASTC. This two-step method never fails.

Example 5 — Allied Angles

Find the values of (a) sin 210° (b) cos 315° (c) tan 120° using allied angle formulas.

Solution: (a) sin 210° = sin(180° + 30°) = −sin 30° = −1/2. (b) cos 315° = cos(360° − 45°) = cos 45° = 1/√2. (c) tan 120° = tan(180° − 60°) = −tan 60° = −√3.

Example 6 — Sign Convention Problem

If sin θ = −1/2 and θ lies in the third quadrant, find cos θ and tan θ without using a calculator.

Solution: In QIII, both cos and sin are negative, while tan is positive. The reference angle whose sine is 1/2 is 30°. Since the angle is in QIII, θ = 180° + 30° = 210°. Therefore cos 210° = −cos 30° = −√3/2, and tan 210° = tan 30° = 1/√3.

2. Trigonometric Equations

Trigonometric equations involve unknown angles that satisfy a given trigonometric relation. NDA frequently tests general solutions and principal values with 2 to 3 questions per paper. The key is to first reduce the given equation to one of the standard forms and then apply the general solution formula.

Principal Values

The principal solution is the solution that lies within the principal value range of the corresponding inverse trigonometric function. For NDA problems, the principal solution is typically sought in the interval [0, 2π). To find the principal solutions:

Step 1: Find the acute reference angle α using the inverse function of the absolute value (e.g., α = sin⁻¹(|k|)).
Step 2: Determine the quadrants in which the given function is positive or negative using the ASTC rule.
Step 3: For each applicable quadrant, express the angle in terms of α. In QI: θ = α. In QII: θ = π − α. In QIII: θ = π + α. In QIV: θ = 2π − α.

General Solutions

The general solution accounts for all possible angles that satisfy the equation by adding integer multiples of the fundamental period of the function. The standard forms are:

Equation	General Solution	Condition / Notes
sin θ = sin α	θ = nπ + (−1)ⁿ α	n ∈ Z, also written as θ = nπ + (−1)ⁿα
cos θ = cos α	θ = 2nπ ± α	n ∈ Z, α ∈ [0, π]
tan θ = tan α	θ = nπ + α	n ∈ Z, α ≠ (2k+1)π/2
sin θ = 0	θ = nπ	n ∈ Z
cos θ = 0	θ = (2n + 1)π/2	n ∈ Z
tan θ = 0	θ = nπ	n ∈ Z
sin θ = 1	θ = (4n + 1)π/2	n ∈ Z
cos θ = 1	θ = 2nπ	n ∈ Z
sin²θ = sin²α	θ = nπ ± α	n ∈ Z
cos²θ = cos²α	θ = nπ ± α	n ∈ Z
tan²θ = tan²α	θ = nπ ± α	n ∈ Z

Systematic Strategy for Solving Trigonometric Equations in NDA

Step 1: Simplify the equation by applying identities to convert it into a single trigonometric function if possible.
Step 2: If the equation is quadratic in sin θ or cos θ, substitute t = sin θ or t = cos θ and solve the quadratic.
Step 3: For each solution of the form sin θ = k or cos θ = k or tan θ = k, identify the reference angle α and determine the quadrants using the sign of k.
Step 4: Write the principal solutions in [0, 2π) by expressing angles in the appropriate quadrants.
Step 5: If the general solution is required, apply the standard formula for the respective trigonometric function.

NDA Shortcut

When solving trig equations in NDA, always find the acute reference angle first using inverse functions. Then use the CAST (ASTC) rule to determine the quadrants where the solutions lie. For general solutions, simply plug into the formula — no need to derive each time. This approach cuts solving time by half.

Example 7 — NDA 2019

Find the general solution of 2 sin²θ − sin θ − 1 = 0.

Solution: Let t = sin θ. Then 2t² − t − 1 = 0. Factoring: (2t + 1)(t − 1) = 0, so t = 1 or t = −1/2.
Case 1: sin θ = 1 → θ = (4n + 1)π/2, n ∈ Z.
Case 2: sin θ = −1/2 = sin(−π/6). Using GS formula: θ = nπ + (−1)ⁿ(−π/6) = nπ + (−1)ⁿ⁺¹π/6, n ∈ Z.

Example 8 — Principal Solutions

Find the principal solutions of tan θ = −√3 in the interval [0, 2π).

Solution: The reference angle α = tan⁻¹(√3) = 60° = π/3. Since tan is negative in QII and QIV: In QII: θ = π − π/3 = 2π/3. In QIV: θ = 2π − π/3 = 5π/3. Therefore the principal solutions are 2π/3 and 5π/3.

Example 9 — Quadratic in Trigonometric Form

Solve 3 cos²θ + 2 sin θ − 3 = 0 for θ ∈ [0, 2π).

Solution: Using cos²θ = 1 − sin²θ: 3(1 − sin²θ) + 2 sin θ − 3 = 0 → −3 sin²θ + 2 sin θ = 0 → sin θ(2 − 3 sin θ) = 0. Therefore sin θ = 0 or sin θ = 2/3. For sin θ = 0: θ = 0, π. For sin θ = 2/3: θ = sin⁻¹(2/3) and π − sin⁻¹(2/3).

NDA Shortcut

For equations involving both sin and cos, always convert to a single function using sin² + cos² = 1. If the equation is of the form a sin θ + b cos θ = c, use the auxiliary method: write it as R sin(θ + φ) where R = √(a² + b²) and φ = tan⁻¹(b/a). This is a powerful NDA trick.

3. Inverse Trigonometric Functions

Inverse trigonometric functions give the angle corresponding to a given trigonometric ratio. Since trigonometric functions are periodic and not one-to-one, their domains are restricted to define the inverse functions uniquely. NDA regularly tests domain, range, properties, and simple evaluations of inverse trigonometric expressions.

Domain and Range

Each inverse function has a restricted principal value branch to ensure that the function is one-to-one and onto. The following table summarises the domain and range (principal value) for each inverse trigonometric function:

Function	Domain	Range (Principal Value Branch)	Nature
sin⁻¹ x	[−1, 1]	[−π/2, π/2]	Increasing
cos⁻¹ x	[−1, 1]	[0, π]	Decreasing
tan⁻¹ x	(−∞, ∞)	(−π/2, π/2)	Increasing
cot⁻¹ x	(−∞, ∞)	(0, π)	Decreasing
sec⁻¹ x	(−∞, −1] ∪ [1, ∞)	[0, π] − {π/2}	Increasing in each branch
cosec⁻¹ x	(−∞, −1] ∪ [1, ∞)	[−π/2, π/2] − {0}	Decreasing in each branch

Properties and Identities

Complementary Relations (Most Frequently Tested in NDA)

For any x in the respective domain, the sum of a function and its co-function equals π/2:

sin⁻¹ x + cos⁻¹ x = π/2 for x ∈ [−1, 1]
tan⁻¹ x + cot⁻¹ x = π/2 for x ∈ R
sec⁻¹ x + cosec⁻¹ x = π/2 for x ∈ (−∞, −1] ∪ [1, ∞)

Negative Arguments

sin⁻¹(−x) = −sin⁻¹ x — odd function
cos⁻¹(−x) = π − cos⁻¹ x — neither even nor odd
tan⁻¹(−x) = −tan⁻¹ x — odd function
cot⁻¹(−x) = π − cot⁻¹ x
sec⁻¹(−x) = π − sec⁻¹ x
cosec⁻¹(−x) = −cosec⁻¹ x

Sum and Difference of Arctangents

tan⁻¹ x + tan⁻¹ y = tan⁻¹[(x + y)/(1 − xy)] — when xy < 1
tan⁻¹ x + tan⁻¹ y = π + tan⁻¹[(x + y)/(1 − xy)] — when xy > 1 and x, y > 0
tan⁻¹ x + tan⁻¹ y = −π + tan⁻¹[(x + y)/(1 − xy)] — when xy > 1 and x, y < 0
tan⁻¹ x − tan⁻¹ y = tan⁻¹[(x − y)/(1 + xy)] — for all x, y

Inverse of Trigonometric Functions

sin⁻¹(sin x) = x — only when x ∈ [−π/2, π/2]
cos⁻¹(cos x) = x — only when x ∈ [0, π]
tan⁻¹(tan x) = x — only when x ∈ (−π/2, π/2)
sin(sin⁻¹ x) = x — for all x ∈ [−1, 1]
cos(cos⁻¹ x) = x — for all x ∈ [−1, 1]
tan(tan⁻¹ x) = x — for all x ∈ R

NDA Shortcut

The complementary relations (sum = π/2) are the most frequently tested inverse trig property in NDA. Also, to evaluate expressions like sin(cos⁻¹ x) or cos(tan⁻¹ x), use the right-triangle method: draw a triangle where the given ratio matches the inner inverse function, then read off the required ratio directly.

Example 10 — NDA 2021

Find the value of tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3).

Solution: tan⁻¹ 1 = π/4. Let S = tan⁻¹ 2 + tan⁻¹ 3. Here xy = 2 × 3 = 6 > 1 and both x, y > 0, so we use the second formula: S = π + tan⁻¹[(2 + 3)/(1 − 6)] = π + tan⁻¹(−1) = π − π/4 = 3π/4. Therefore total = π/4 + 3π/4 = π. This is a classic NDA problem.

Example 11 — Domain of Inverse Functions

Find the domain of f(x) = sin⁻¹(2x − 1) + cos⁻¹(x/2).

Solution: For sin⁻¹(2x − 1): domain requires −1 ≤ 2x − 1 ≤ 1. Adding 1: 0 ≤ 2x ≤ 2. Dividing by 2: 0 ≤ x ≤ 1. For cos⁻¹(x/2): domain requires −1 ≤ x/2 ≤ 1. Multiplying by 2: −2 ≤ x ≤ 2. The domain of f(x) is the intersection of the two domains: [0, 1].

Example 12 — Triangle Method

Evaluate sin(cos⁻¹(3/5)).

Solution: Let α = cos⁻¹(3/5). Then cos α = 3/5, and α ∈ [0, π]. Using the triangle method: draw a right triangle with adjacent = 3 and hypotenuse = 5. By Pythagoras, opposite = √(25 − 9) = √16 = 4. Since α is in [0, π], sin α is non-negative. Therefore sin(cos⁻¹(3/5)) = sin α = opposite/hypotenuse = 4/5.

Example 13 — Inverse Property

Find the value of sin⁻¹(sin(5π/6)).

Solution: We cannot directly say sin⁻¹(sin(5π/6)) = 5π/6 because 5π/6 ∉ [−π/2, π/2]. First, sin(5π/6) = sin(π − π/6) = sin(π/6) = 1/2. Then sin⁻¹(1/2) = π/6 (principal value). Therefore sin⁻¹(sin(5π/6)) = π/6.

NDA Shortcut

Remember: sin⁻¹(sin x) = x only when x is in [−π/2, π/2]. Similarly cos⁻¹(cos x) = x only when x is in [0, π]. NDA often tests this property by giving angles outside these ranges. The trick is to first reduce the angle to an equivalent angle within the principal range using allied angle formulas.

4. Heights and Distances

Heights and distances problems use trigonometric ratios to measure inaccessible heights and distances without direct measurement. This is one of the most practical applications of trigonometry and appears regularly in NDA Paper 1 with 1 to 2 questions. These problems typically involve angles of elevation, depression, and bearings.

Angle of Elevation and Depression

Angle of Elevation: When an observer looks up at an object, the angle between the horizontal line of sight and the line joining the observer's eye to the object is called the angle of elevation.

Angle of Depression: When an observer looks down at an object, the angle between the horizontal line of sight and the line joining the observer's eye to the object is called the angle of depression.

The fundamental relationship in all such problems is: tan(angle) = opposite/adjacent = height/distance. In a right triangle formed by the observer, the base, and the object: tan θ = h / d, where h is the height (vertical difference) and d is the horizontal distance.

Application Problems

The standard types of problems that appear in NDA include:

Single angle problems: Given the angle of elevation/depression and one side, find the other side using tan θ.
Two-angle problems: The observer moves towards or away from the object, giving two different angles. Set up two equations and solve for the unknown height or distance.
Problems with two objects: From a high point, two objects are observed at different angles of depression. Find the distance between them.
Bearing problems: Directions are given as bearings (angles measured from north). These require combining trigonometry with geometry.

General strategy for solving heights and distances problems:

Step 1: Draw a clear diagram representing the situation. Label all known and unknown quantities.
Step 2: Identify the right triangles formed. For each triangle, write the tangent ratio: tan(angle) = opposite/adjacent.
Step 3: Set up equations based on the given information. If there are two unknowns, you generally need two equations.
Step 4: Solve the equations simultaneously. Use standard trigonometric values for common angles.
Step 5: Check if the answer is reasonable and in the correct units.

Example 14 — NDA 2020

From the top of a 75 m high lighthouse, the angles of depression of two ships are 30° and 45° respectively. If both ships are on the same side of the lighthouse, find the distance between the two ships.

Solution: Let the lighthouse base be point O, and the ships be at points A (nearer) and B (farther). Let OA = d₁, OB = d₂. From the top, angle of depression of the nearer ship is 45°, so tan 45° = 75/d₁ → 1 = 75/d₁ → d₁ = 75 m. For the farther ship, angle of depression is 30°, so tan 30° = 75/d₂ → 1/√3 = 75/d₂ → d₂ = 75√3 m. The distance between the ships = d₂ − d₁ = 75(√3 − 1) m ≈ 75(1.732 − 1) = 75 × 0.732 = 54.9 m.

Example 15 — Moving Object Problem

A man on the top of a tower observes a car approaching the tower at a uniform speed. When the car is at point P, the angle of depression is 30°. After 6 seconds, when the car reaches point Q, the angle of depression becomes 60°. Find the total time taken by the car to reach the tower from point P.

Solution: Let the tower height be h. At P (initial position): d₁ = h cot 30° = h√3. At Q (after 6 sec): d₂ = h cot 60° = h/√3. Distance covered in 6 seconds = d₁ − d₂ = h(√3 − 1/√3) = h(2/√3). Remaining distance from Q to the tower = d₂ = h/√3. Since speed is constant: time remaining = 6 × (h/√3) / (h(2/√3)) = 6 × (1/2) = 3 seconds. Total time from P to tower = 6 + 3 = 9 seconds.

NDA Shortcut

For moving object problems in heights and distances, the distance covered in a given time interval is proportional to (cot θ₁ − cot θ₂), where θ₁ and θ₂ are the angles of depression/elevation. Set up a direct proportion to find unknown time without computing the height explicitly. This saves precious minutes in the exam.

Example 16 — Two Positions, Same Object

From a point on the ground, the angle of elevation of the top of a tower is 30°. After walking 40 metres towards the tower, the angle of elevation becomes 60°. Find the height of the tower.

Solution: Let the height of the tower be h metres. Let the distance from the second observation point to the base of the tower be x metres. From the first position (distance = x + 40): tan 30° = h/(x + 40) → 1/√3 = h/(x + 40) → h = (x + 40)/√3. From the second position (distance = x): tan 60° = h/x → √3 = h/x → h = √3 x. Equating: (x + 40)/√3 = √3 x → x + 40 = 3x → 2x = 40 → x = 20 m. Therefore h = √3 × 20 = 20√3 metres.

NDA Shortcut

In two-position problems (where the observer moves), always let the distance from the second (nearer) position to the base be x. Then express the height h in two ways using tan of each angle. Equate the two expressions to solve for x, then find h. This approach avoids dealing with two unknown variables simultaneously and reduces algebraic errors.

5. Trigonometric Formulas and Transformations

Formula-based problems form the backbone of trigonometry in NDA. Compound angles, multiple angles, and sum-to-product identities are heavily tested. Students must be able to apply these formulas fluently to simplify expressions, prove identities, and solve equations.

Compound Angle Formulas

Compound angle formulas express the trigonometric functions of sums or differences of two angles in terms of functions of the individual angles:

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
tan(A + B) = (tan A + tan B) / (1 − tan A tan B)
tan(A − B) = (tan A − tan B) / (1 + tan A tan B)
cot(A + B) = (cot A cot B − 1) / (cot B + cot A)
cot(A − B) = (cot A cot B + 1) / (cot B − cot A)

Multiple and Submultiple Angle Formulas

Double Angle Formulas

sin 2A = 2 sin A cos A = 2 tan A / (1 + tan² A)
cos 2A = cos² A − sin² A = 2 cos² A − 1 = 1 − 2 sin² A = (1 − tan² A) / (1 + tan² A)
tan 2A = 2 tan A / (1 − tan² A)

Triple Angle Formulas

sin 3A = 3 sin A − 4 sin³ A
cos 3A = 4 cos³ A − 3 cos A
tan 3A = (3 tan A − tan³ A) / (1 − 3 tan² A)

Submultiple Angle Formulas

sin A = 2 sin(A/2) cos(A/2) — derived from sin 2θ = 2 sin θ cos θ with θ = A/2
cos A = cos²(A/2) − sin²(A/2) = 2 cos²(A/2) − 1 = 1 − 2 sin²(A/2)
tan A = 2 tan(A/2) / (1 − tan²(A/2))
sin A = 2 tan(A/2) / (1 + tan²(A/2)) — t-formula
cos A = (1 − tan²(A/2)) / (1 + tan²(A/2)) — t-formula

Sum-to-Product and Product-to-Sum Formulas

Product-to-Sum Formulas

These formulas convert products of trigonometric functions into sums or differences. They are especially useful for evaluating products of cosines and for integration:

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)

Sum-to-Product Formulas

These formulas convert sums or differences of trigonometric functions into products. They are useful for solving equations and simplifying expressions:

sin C + sin D = 2 sin[(C + D)/2] cos[(C − D)/2]
sin C − sin D = 2 cos[(C + D)/2] sin[(C − D)/2]
cos C + cos D = 2 cos[(C + D)/2] cos[(C − D)/2]
cos C − cos D = −2 sin[(C + D)/2] sin[(C − D)/2]

Comprehensive Formula Reference Table

Category	Formula	Typical NDA Application
Compound	sin(A + B) = sin A cos B + cos A sin B	Evaluating sin 75°, cos 15° etc.
Compound	cos(A + B) = cos A cos B − sin A sin B	Finding exact trigonometric values
Compound	tan(A + B) = (tan A + tan B)/(1 − tan A tan B)	Proof of π/4 + tan⁻¹x type problems
Double Angle	sin 2A = 2 sin A cos A	Simplifying product expressions
Double Angle	cos 2A = 2 cos² A − 1 = 1 − 2 sin² A	Converting between powers of sin and cos
Double Angle	tan 2A = 2 tan A / (1 − tan² A)	Trigonometric equations
Triple Angle	sin 3A = 3 sin A − 4 sin³ A	Cubic trigonometric equations
Triple Angle	cos 3A = 4 cos³ A − 3 cos A	Identity proof problems
Product→Sum	2 sin A cos B = sin(A+B) + sin(A−B)	Evaluating products of sines and cosines
Product→Sum	2 cos A cos B = cos(A+B) + cos(A−B)	Converting product to sum for integration
Sum→Product	sin C + sin D = 2 sin((C+D)/2) cos((C−D)/2)	Solving equations with sums of trig functions
Sum→Product	cos C − cos D = −2 sin((C+D)/2) sin((C−D)/2)	Simplifying difference of cosines

NDA Shortcut

The double-angle formula cos 2A = 2 cos²A − 1 = 1 − 2 sin²A is extremely useful for converting between squared and linear trigonometric terms. Also, the product-to-sum trick (multiply and divide by 2 sin of the smallest angle) solves problems like cos 20° cos 40° cos 80° in just a few steps.

Example 17 — Product-to-Sum (Classic NDA Problem)

Find the value of cos 20° cos 40° cos 80°.

Solution: Multiply numerator and denominator by 2 sin 20°: (2 sin 20° cos 20°) cos 40° cos 80° / (2 sin 20°) = (sin 40° cos 40° cos 80°) / (2 sin 20°) = (sin 80° cos 80°) / (4 sin 20°) = sin 160° / (8 sin 20°) = sin(180° − 20°) / (8 sin 20°) = sin 20° / (8 sin 20°) = 1/8. This technique is essential for NDA.

Example 18 — Compound Angle Application

If sin A = 3/5 and cos B = 5/13, with A and B both acute, find (a) sin(A + B) and (b) tan(A − B).

Solution: First find cos A = √(1 − 9/25) = 4/5. sin B = √(1 − 25/169) = 12/13.
(a) sin(A+B) = sinA cosB + cosA sinB = (3/5)(5/13) + (4/5)(12/13) = 15/65 + 48/65 = 63/65.
(b) tan A = 3/4, tan B = 5/12. tan(A−B) = (tanA − tanB)/(1 + tanA tanB) = (3/4 − 5/12)/(1 + (3/4)(5/12)) = (9/12 − 5/12)/(1 + 15/48) = (4/12)/(63/48) = (1/3)/(21/16) = 16/63.

Example 19 — Sum-to-Product

Express sin 7θ + sin 5θ as a product and find its value when θ = π/12.

Solution: Using sin C + sin D = 2 sin((C+D)/2) cos((C−D)/2): sin 7θ + sin 5θ = 2 sin(6θ) cos(θ). At θ = π/12: 2 sin(π/2) cos(π/12) = 2(1) cos(π/12) = 2 cos(π/12). Since cos(π/12) = cos 15° = (√6 + √2)/4, the value is 2 × (√6 + √2)/4 = (√6 + √2)/2.

NDA Shortcut

When applying sum-to-product formulas, the average (C + D)/2 and half-difference (C − D)/2 follow naturally from the original angles. A common NDA trap: the formula for cos C − cos D includes a negative sign, which many students forget. Always double-check the sign in this formula.

Key Formula Summary for NDA Trigonometry

Pythagorean Identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ.
Compound Angles: 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).
Double Angle: sin 2A = 2 sin A cos A; cos 2A = 2 cos²A − 1 = 1 − 2 sin²A; tan 2A = 2 tan A/(1 − tan²A).
Triple Angle: sin 3A = 3 sin A − 4 sin³A; cos 3A = 4 cos³A − 3 cos A.
Product to Sum: 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).
Sum to Product: 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); cos C − cos D = −2 sin((C+D)/2) sin((C−D)/2).
General Solutions: sin θ = sin α → θ = nπ + (−1)ⁿα; cos θ = cos α → θ = 2nπ ± α; tan θ = tan α → θ = nπ + α.
Inverse Trig Properties: sin⁻¹x + cos⁻¹x = π/2; tan⁻¹x + cot⁻¹x = π/2; tan⁻¹x + tan⁻¹y formula with xy condition.
Heights and Distances: tan θ = height/distance; always draw a clear diagram first; use cot for distance when height is known.

Practice Questions

← Calculus Coordinate Geometry →