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.

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:

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

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

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

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

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

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

Triple Angle Formulas

The triple angle identities express trigonometric ratios of $3A$ in terms of $A$:

$\sin 3A = 3\sin A - 4\sin^3 A$

$\cos 3A = 4\cos^3 A - 3\cos A$

$\tan 3A = \dfrac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$

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: