Trigonometric Ratios of Compound Angles

Compound Angles in Trigonometry

In trigonometry, a compound angle is formed when two or more angles are added or subtracted. If $A$, $B$, and $C$ are angles, then expressions such as $A + B$, $A - B$, $A + B + C$, and $A + B - C$ are all called compound angles.

The study of compound angles is important because many trigonometric values cannot be evaluated directly using standard angles. This leads to the use of trigonometric ratios of compound angles, which are widely used in simplification, proofs, and problem-solving.

Basics of Compound Angles

Before learning compound angle formulas, it is important to understand what compound angles actually are. Compound angles are formed when two angles are added or subtracted, resulting in expressions such as $(A+B)$ or $(A-B)$. These expressions are widely used in trigonometry to derive identities, simplify calculations, and solve advanced mathematical problems. A clear understanding of angle sums, angle differences, and basic trigonometric ratios makes it easier to apply compound angle formulas correctly.

What is an Angle Sum?

An angle sum is formed when two angles are added together to create a new angle. In trigonometry, angle sums are commonly represented as $(A+B)$ and are used in formulas involving sine, cosine, and tangent functions.

For example:

• $30^\circ+60^\circ=90^\circ$
• $45^\circ+45^\circ=90^\circ$
• $15^\circ+30^\circ=45^\circ$

When dealing with trigonometric functions, the value of $\sin(A+B)$ is not simply equal to $\sin A+\sin B$. Instead, special compound angle formulas are required to evaluate such expressions accurately.

Angle sums are widely used in:

• Trigonometric identities
• Coordinate geometry
• Calculus
• Physics and engineering applications
• Navigation and surveying calculations

What is an Angle Difference?

An angle difference is obtained when one angle is subtracted from another angle. It is represented by expressions such as $(A-B)$.

Examples include:

• $60^\circ-30^\circ=30^\circ$
• $90^\circ-45^\circ=45^\circ$
• $120^\circ-60^\circ=60^\circ$

Just like angle sums, trigonometric functions involving angle differences require special formulas.

For example:

• $\sin(A-B)$
• $\cos(A-B)$
• $\tan(A-B)$

These formulas help evaluate trigonometric ratios of angles that may not appear directly in standard trigonometric tables.

Angle difference formulas are particularly useful when:

• Simplifying trigonometric expressions
• Solving equations
• Finding exact values of angles
• Proving trigonometric identities

Understanding Compound Angle Expressions

A compound angle expression contains two angles connected by either addition or subtraction.

General forms include:

• $(A+B)$
• $(A-B)$
• $(\alpha+\beta)$
• $(\theta+\phi)$

Some common trigonometric compound angle expressions are:

• $\sin(A+B)$
• $\sin(A-B)$
• $\cos(A+B)$
• $\cos(A-B)$
• $\tan(A+B)$
• $\tan(A-B)$

These expressions cannot be evaluated by simply applying arithmetic operations to individual trigonometric values. Instead, they follow specific identities known as compound angle formulas.

For example:

$\sin(30^\circ+60^\circ)$ cannot be calculated as: $\sin30^\circ+\sin60^\circ$

Instead, the compound angle formula must be used.

Understanding these expressions is essential because they serve as the foundation for:

• Double angle formulas
• Triple angle formulas
• Multiple angle identities
• Advanced trigonometric proofs

Review of Basic Trigonometric Ratios

Before studying compound angles, students should be familiar with the six basic trigonometric ratios.

For a right-angled triangle:

Important relationships include:

• $\sin^2\theta+\cos^2\theta=1$
• $\tan\theta=\frac{\sin\theta}{\cos\theta}$
• $\sec^2\theta=1+\tan^2\theta$
• $\csc^2\theta=1+\cot^2\theta$

These fundamental identities are frequently used while deriving and applying compound angle formulas.

Properties of Compound Angle Formulas

Compound angle formulas possess several important mathematical properties that make them useful for simplifying expressions and solving trigonometric problems. Understanding these properties helps students recognize patterns and apply formulas more effectively in examinations and advanced mathematics.

Symmetry Properties

Many compound angle formulas exhibit symmetry, meaning their structure follows predictable mathematical patterns.

For sine:

• $\sin(A+B)=\sin A\cos B+\cos A\sin B$
• $\sin(B+A)=\sin B\cos A+\cos B\sin A$

Since addition is commutative, $\sin(A+B)=\sin(B+A)$

For cosine:

• $\cos(A+B)=\cos A\cos B-\sin A\sin B$
• $\cos(B+A)=\cos B\cos A-\sin B\sin A$

Therefore, $\cos(A+B)=\cos(B+A)$

These symmetry properties make many trigonometric manipulations easier.

Important observations:

• Sine and cosine formulas remain unchanged when the order of angles is interchanged.
• Symmetry helps verify answers quickly.
• Many trigonometric identities are derived using symmetry principles.

Periodicity Properties

Trigonometric functions repeat their values after specific intervals. This repeating behavior is known as periodicity.

The periods of common trigonometric functions are:

Examples:

• $\sin(\theta+2\pi)=\sin\theta$
• $\cos(\theta+2\pi)=\cos\theta$
• $\tan(\theta+\pi)=\tan\theta$

Because of periodicity, compound angle formulas remain valid even when large multiples of $\pi$ or $360^\circ$ are added to the angles.

Applications include:

• Simplifying large-angle expressions
• Solving trigonometric equations
• Wave and oscillation analysis
• Signal processing

Sign Conventions

The sign of a trigonometric function depends on the quadrant in which the angle lies.

The standard quadrant rule is often remembered using:

ASTC (All Students Take Calculus)

Examples:

• $\sin150^\circ$ is positive.
• $\cos150^\circ$ is negative.
• $\tan210^\circ$ is positive.
• $\cos330^\circ$ is positive.

Sign conventions become extremely important when applying compound angle formulas because an incorrect sign often leads to an incorrect final answer.

Students should always:

• Identify the quadrant first.
• Determine the sign of each trigonometric ratio.
• Apply the compound angle formula carefully.

Relationship Between Different Trigonometric Functions

Compound angle formulas reveal strong relationships between various trigonometric functions.

Some important relationships are:

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

$\cot\theta=\frac{\cos\theta}{\sin\theta}$

$\sec\theta=\frac{1}{\cos\theta}$

$\csc\theta=\frac{1}{\sin\theta}$

Using the sine and cosine compound angle formulas, the tangent compound angle formula can be derived.

For example:

$\tan(A+B)=\frac{\sin(A+B)}{\cos(A+B)}$

Substituting the compound angle identities for sine and cosine leads directly to the tangent addition formula.

This interconnected nature of trigonometric functions allows mathematicians to:

• Derive new identities
• Simplify expressions
• Solve equations efficiently
• Develop advanced trigonometric formulas

Understanding these relationships is essential for mastering trigonometric ratios of compound angles and forms the basis for higher-level trigonometry, calculus, and mathematical analysis.

Trigonometric Ratios of Compound Angles

The trigonometric ratios of compound angles express the sine, cosine, and tangent of the sum or difference of two angles in terms of the trigonometric ratios of the individual angles. These identities are also known as sum and difference formulas.

Sine of Compound Angles

$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

Cosine of Compound Angles

$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

These formulas are frequently used to find exact values of trigonometric expressions and to prove identities.

Tangent Compound Angle Formulas

Tangent of Sum and Difference of Angles

$\tan(\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

$\tan(\alpha - \beta) = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

These identities are especially useful in algebraic simplifications and equation solving.

Cotangent Compound Angle Formulas

Cotangent of Sum and Difference of Angles

$\cot(\alpha + \beta) = \dfrac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta}$

$\cot(\alpha - \beta) = \dfrac{\cot \alpha \cot \beta + 1}{\cot \beta - \cot \alpha}$

Cotangent formulas are often applied in advanced trigonometric proofs and transformations.

Derivations of Compound Angle Formulas

The derivations of compound angle formulas help us understand why formulas like $\sin(\alpha \pm \beta)$ and $\cos(\alpha \pm \beta)$ work, rather than just memorising them. These derivations are commonly based on unit circle geometry, coordinate geometry, and triangle properties, making them important for conceptual clarity in trigonometry.

Derivation of $\cos(\alpha - \beta)$ Using the Unit Circle

To derive the cosine of the difference of two angles, we use the unit circle and coordinate geometry approach, which is both intuitive and mathematically sound.

Step 1: Represent Angles on the Unit Circle

Consider a unit circle centered at the origin $O$.

• Let point $P$ lie on the unit circle at an angle $\alpha$ from the positive $x$-axis.
  Its coordinates are $(\cos \alpha, \sin \alpha)$.

• Let point $Q$ lie on the unit circle at an angle $\beta$ from the positive $x$-axis.
  Its coordinates are $(\cos \beta, \sin \beta)$.

The angle between the rays $OP$ and $OQ$ is therefore $\alpha - \beta$.

Step 2: Introduce a Reference Triangle

Now consider two additional points:

• Point $A$ lies on the unit circle at an angle $(\alpha - \beta)$ from the positive $x$-axis, with coordinates
  $(\cos(\alpha - \beta), \sin(\alpha - \beta))$.

• Point $B$ lies on the positive $x$-axis at coordinates $(1, 0)$.

Triangle $POQ$ can be viewed as a rotation of triangle $AOB$, since both triangles subtend the same angle at the center.

Step 3: Use Distance Equality

Because rotation does not change distances:

• The distance between points $P$ and $Q$

• is equal to the distance between points $A$ and $B$

Using the distance formula, we can equate these two distances.

Step 4: Apply the Distance Formula

Distance between $P(\cos \alpha, \sin \alpha)$ and $Q(\cos \beta, \sin \beta)$:

$(PQ)^2 = (\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2$

Distance between $A(\cos(\alpha - \beta), \sin(\alpha - \beta))$ and $B(1, 0)$:

$(AB)^2 = (\cos(\alpha - \beta) - 1)^2 + \sin^2(\alpha - \beta)$

Since $PQ = AB$, equating these expressions and simplifying leads to the identity:

$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

Final Result

The cosine of the difference of two angles is given by:

$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

Derivation of $\cos(\alpha + \beta)$

Using the identity $\cos(\alpha + \beta) = \cos(\alpha - (-\beta))$,

$\cos(\alpha + \beta) = \cos \alpha \cos(-\beta) + \sin \alpha \sin(-\beta)$

Using even–odd properties,

$\cos(-\beta) = \cos \beta$

$\sin(-\beta) = -\sin \beta$

So, $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

Sine Compound Angle Formulas

Derivation of $\sin(\alpha - \beta)$

$\sin(\alpha - \beta) = \cos\left(90^\circ - (\alpha - \beta)\right)$

$= \cos\left((90^\circ - \alpha) + \beta\right)$

$= \cos(90^\circ - \alpha)\cos \beta - \sin(90^\circ - \alpha)\sin \beta$

$= \sin \alpha \cos \beta - \cos \alpha \sin \beta$

Derivation of $\sin(\alpha + \beta)$

$\sin(\alpha + \beta) = \sin(\alpha - (-\beta))$

$= \sin \alpha \cos(-\beta) - \cos \alpha \sin(-\beta)$

$= \sin \alpha \cos \beta + \cos \alpha \sin \beta$

Tangent Compound Angle Formulas

Derivation of $\tan(\alpha + \beta)$

$\tan(\alpha + \beta) = \dfrac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)}$

$= \dfrac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta}$

Dividing numerator and denominator by $\cos \alpha \cos \beta$,

$\tan(\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

Tangent Difference Formula

$\tan(\alpha - \beta) = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

Cotangent Compound Angle Formulas

Derivation of $\cot(\alpha + \beta)$

$\cot(\alpha + \beta) = \dfrac{\cos(\alpha + \beta)}{\sin(\alpha + \beta)}$

$= \dfrac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\sin \alpha \cos \beta + \cos \alpha \sin \beta}$

Dividing numerator and denominator by $\sin \alpha \sin \beta$,

$\cot(\alpha + \beta) = \dfrac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta}$

Cotangent Difference Formula

$\cot(\alpha - \beta) = \dfrac{\cot \alpha \cot \beta + 1}{\cot \beta - \cot \alpha}$

Important Results Using Compound Angles

$\sin(A+B)\sin(A-B) = \sin^2 A - \sin^2 B = \cos^2 B - \cos^2 A$

$\cos(A+B)\cos(A-B) = \cos^2 A - \sin^2 B = \cos^2 B - \sin^2 A$

$\tan(A+B+C) = \dfrac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan B \tan C - \tan C \tan A}$

Transformation of Products into Sum or Difference of Trigonometric Functions

In trigonometry, product-to-sum formulas help convert the product of sine and cosine functions into a sum or difference of trigonometric ratios. These identities are extremely useful for simplifying expressions, proving identities, and solving trigonometric equations, especially in board and competitive exams.

Product-to-Sum Formulas

The standard transformation of products into sums or differences is given below:

$2\sin A \cos B = \sin(A+B) + \sin(A-B)$

$2\cos A \cos B = \cos(A+B) + \cos(A-B)$

$2\cos A \sin B = \sin(A+B) - \sin(A-B)$

$2\sin A \sin B = \cos(A-B) - \cos(A+B)$

These formulas allow us to rewrite products of trigonometric functions in a simpler and more workable form.

Multiple and Submultiple Angle Formulas

Multiple-angle identities express trigonometric ratios of angles like $2A$ or $3A$ in terms of the ratio of angle $A$. These identities play a key role in simplification, evaluation, and solving higher-level trigonometric problems.

Double Angle Formulas

The double angle identities are derived from compound angle formulas:

$\sin 2A = 2\sin A \cos A$

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

$\cos 2A = \cos^2 A - \sin^2 A$

$\cos 2A = 2\cos^2 A - 1$

$\cos 2A = 1 - 2\sin^2 A$

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

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

Best Books for Trigonometric Ratios of Compound Angles

A strong understanding of compound angle formulas is essential for mastering trigonometric identities, equations, inverse trigonometry, and calculus. The following books provide detailed theory, derivations, and practice questions.

Shortcut Tips and Tricks for Compound Angle Questions

Remembering a few standard patterns can make solving compound angle problems much faster and reduce calculation errors.

Important Formula Table

This formula table summarizes the most important compound angle identities used in trigonometry.

Solved Examples Based on Trigonometric ratios of compound angles

Example 1: The value of $\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}$ is
[JEE MAINS 2023]

1. 4
2. 3
3. 2
4. 1

Solution

$\left(\tan 9^{\circ}+\cot 9^{\circ}\right)-\left(\tan 27^{\circ}+\cot 27^{\circ}\right)$

$\dfrac{1}{\sin 9^{\circ} \cos 9^{\circ}}-\dfrac{1}{\sin 27^{\circ} \cos 27^{\circ}}$

$\dfrac{2}{\sin 18^{\circ}}-\dfrac{2}{\sin 54^{\circ}}$

$\dfrac{2(4)}{\sqrt{5}-1}-\dfrac{2(4)}{\sqrt{5}+1}$

$\dfrac{8(\sqrt{5}+1)}{4}-\dfrac{8(\sqrt{5}-1)}{4}$

$2[(\sqrt{5}+1)-(\sqrt{5}-1)]$

$=4$

Hence, the answer is option 4.

Example 2: Let $\mathrm{S}={\theta\in[0,2\pi]:8^{2\sin^2\theta}+8^{2\cos^2\theta}=16}$

Then $\mathrm{n(S)}+\sum_{\theta\in S}\left(\sec\left(\dfrac{\pi}{4}+2\theta\right)\csc\left(\dfrac{\pi}{4}+2\theta\right)\right)$ equals
[JEE MAINS 2022]

1. 0
2. −2
3. −4
4. 12

Solution

$8^{2\sin^2\theta}+8^{2-2\sin^2\theta}=16$

$y+\dfrac{64}{y}=16$

$y=8$

$\sin^2\theta=\dfrac{1}{2}$

So, $\mathrm{n(S)}+\sum_{\theta\in S}\dfrac{1}{\cos\left(\dfrac{\pi}{4}+2\theta\right)\sin\left(\dfrac{\pi}{4}+2\theta\right)}$

$=4+(-2)\times4$

$=-4$

Hence, the answer is option 3.

Example 3: If $\cot\alpha=1$ and $\sec\beta=-\dfrac{5}{3}$, where
$\pi<\alpha<\dfrac{3\pi}{2}$ and $\dfrac{\pi}{2}<\beta<\pi$, find $\tan(\alpha+\beta)$ and the quadrant.
[JEE MAINS 2022]

1. $-\dfrac{1}{7}$ and IV quadrant
2. $7$ and I quadrant
3. $-7$ and IV quadrant
4. $\dfrac{1}{7}$ and I quadrant

Solution

$\cot\alpha=1 \Rightarrow \tan\alpha=1$

$\sec\beta=-\dfrac{5}{3} \Rightarrow \tan\beta=-\dfrac{4}{3}$

$\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

$=\dfrac{1-\dfrac{4}{3}}{1+\dfrac{4}{3}}$

$=-\dfrac{1}{7}$

$\pi<\alpha<\dfrac{3\pi}{2}$ and $\dfrac{\pi}{2}<\beta<\pi$

$\Rightarrow \dfrac{3\pi}{2}<\alpha+\beta<\dfrac{5\pi}{2}$

$\tan(\alpha+\beta)<0 \Rightarrow$ IV quadrant

Hence, the answer is option 1.

Example 4: Let $f(x+y)+f(x-y)=2f(x)f(y)$ and $f\left(\dfrac{1}{2}\right)=-1$
[JEE MAINS 2021]

Solution

$f(x+y)+f(x-y)=2f(x)f(y)$

Clearly, $f(x)=\cos(ax)$

$f\left(\dfrac{1}{2}\right)=-1$

$-1=\cos\left(-\dfrac{a}{2}\right)$

$-\dfrac{a}{2}=\pi \Rightarrow a=-2\pi$

$f(x)=\cos(-2\pi x)$

For $k\in\mathbb{N}$, $f(k)=1$

$\dfrac{1}{\sin k\sin(k+1)}$

$=\dfrac{1}{\sin1}\left(\cot k-\cot(k+1)\right)$

Required sum

$=\dfrac{1}{\sin1}(\cot1-\cot21)$

$=\dfrac{1}{\sin1}\left(\dfrac{\cos1}{\sin1}-\dfrac{\cos21}{\sin21}\right)$

$=\dfrac{\sin(21-1)}{\sin^2 1\cdot\sin21}$

$=\csc^2(1)\csc(21)\sin(20)$

Hence, the answer is option 3.

List of Topics Related to the Trigonometric Ratios of Compound Angles

This section outlines all important and related topics connected to trigonometric ratios of compound angles, helping you build strong conceptual links. It serves as a quick reference guide for revision, practice, and exam-oriented preparation.

NCERT Resources

We have given below the list of NCERT resources, including NCERT Notes, solutions and exemplar solutions for class 11, chapter 3 Trigonometric Functions:

NCERT Class 11 Chapter 3 Trigonometric Functions Notes

NCERT Class 11 Chapter 3 Trigonometric Functions Solutions

NCERT Exemplar Class 11 Chapter 3 Trigonometric Functions

Practice Questions on Trigonometric Ratios of Compound Angles

This section includes carefully selected practice questions on trigonometric ratios of compound angles, designed to strengthen your understanding of sum and difference formulas. These problems help you apply compound angle identities effectively in board exams and competitive exams like JEE.

Trigonometric Ratios Of Compound Angles - Practice Question

We have provided the list of practice questions based on the following topics: