Sum to Product Formula: List, Proof, Examples, Application

Sum to Product Formula (Sum/Difference into Product)

The sum to product formula (also called sum/difference into product) is an important concept in trigonometry used to convert the sum or difference of sine and cosine functions into a product of trigonometric functions. These formulas are extremely useful for simplifying trigonometric expressions, solving equations, and handling problems in board exams and competitive exams like JEE.

What Is the Sum/Difference to Product Formula?

The sum and difference to product formulas allow us to rewrite expressions involving
$\sin \alpha \pm \sin \beta$ and $\cos \alpha \pm \cos \beta$
as products of sine and cosine terms. This transformation reduces complexity and makes further calculations easier.

Here, $\alpha$ and $\beta$ are any two angles.

List of Sum to Product Formulas

We have given below the list of sum-to-product formulas:

Sum to Product Formulas for Sine

$\sin \alpha + \sin \beta = 2 \sin \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha - \beta}{2}\right)$

$\sin \alpha - \sin \beta = 2 \sin \left(\frac{\alpha - \beta}{2}\right) \cos \left(\frac{\alpha + \beta}{2}\right)$

Sum to Product Formulas for Cosine

$\cos \alpha + \cos \beta = 2 \cos \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha - \beta}{2}\right)$

$\cos \alpha - \cos \beta = -2 \sin \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha - \beta}{2}\right)$

Interpretation of Sum to Product Formulas

• These formulas convert sums or differences of trigonometric functions into products

• They are derived using angle addition and subtraction identities

• They help simplify expressions involving multiple trigonometric terms

• Widely used in trigonometric simplification, integration, and equation solving

Proof of Sum to Product Formula (Using Substitution)

We start from the product to sum identity:

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

Let:

$\alpha = \frac{u + v}{2}$
$\beta = \frac{u - v}{2}$

Then:

$\alpha + \beta = u$
$\alpha - \beta = v$

Substituting into the identity:

$2 \sin \left(\frac{u + v}{2}\right) \cos \left(\frac{u - v}{2}\right) = \sin u + \sin v$

This proves the sum to product identity for sine.
All other sum to product formulas can be derived in a similar manner using standard trigonometric identities.

Important Notes on Sum to Product Formulas

• Sum to product formulas express addition or subtraction as multiplication

• They are derived from product to sum identities

• Helpful in simplifying trigonometric expressions and equations

• Frequently used in Class 11–12 trigonometry and competitive exams

Summary of Key Sum to Product Identities

$\sin A + \sin B = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$

$\sin A - \sin B = 2 \sin \left(\frac{A - B}{2}\right) \cos \left(\frac{A + B}{2}\right)$

$\cos A + \cos B = 2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$

$\cos A - \cos B = -2 \sin \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$

Relationship Between Sum to Product and Product to Sum Formulas

The sum to product formulas and product to sum formulas are closely connected and work as inverse tools in trigonometry. While product to sum identities convert a product of trigonometric functions into a sum or difference, sum to product identities do the exact opposite—they transform a sum or difference into a product.

• Product to Sum: converts expressions like $\sin A \cos B$ into $\sin(A+B)$ and $\sin(A-B)$

• Sum to Product: converts expressions like $\sin A + \sin B$ into $2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$

This inverse relationship helps in:

• Simplifying long trigonometric expressions

• Solving trigonometric equations efficiently

• Converting expressions into forms suitable for integration and series problems

Understanding how these two sets of formulas complement each other is essential for Class 11 trigonometry, NCERT problems, and competitive exams like JEE.

Key Trigonometric Identities Used in Sum to Product Formulas

The derivation and application of sum to product formulas rely on a few fundamental trigonometric identities. These identities form the backbone of most transformations in trigonometry.

• Sum and Difference Identities
  $\sin(A+B)$, $\sin(A-B)$, $\cos(A+B)$, and $\cos(A-B)$ are directly used to derive sum to product formulas.

• Pythagorean Identity
  $\sin^2 x + \cos^2 x = 1$
  Helps in simplification after conversion.

• Even and Odd Function Properties
  $\sin(-x) = -\sin x$, $\cos(-x) = \cos x$
  Useful when handling negative angles during transformation.

• Angle Averaging Concept
  The terms $\frac{A+B}{2}$ and $\frac{A-B}{2}$ naturally arise from rearranging sum and difference identities.

Mastery of these identities ensures error-free application of sum to product trigonometric formulas.

Applications of Sum to Product Formula

The sum to product formula has wide-ranging applications in both academic mathematics and applied problem-solving.

• Simplifying trigonometric expressions involving sums or differences

• Solving trigonometric equations with multiple angles

• Evaluating trigonometric sums quickly in competitive exams

• Converting expressions into integrable forms in calculus

• Handling wave equations and oscillatory motion in physics

Solved Examples Based on Sum/Difference into Product

Example 1: The value of $\cos 75^{\circ}+\cos 15^{\circ}$ is.

Solution

Given that, $\cos 75^{\circ}+\cos 15^{\circ}$

Using the trigonometric formula,
$\cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$

Therefore,
$\cos 75^{\circ}+\cos 15^{\circ}=2 \cos \left(\frac{75^{\circ}+15^{\circ}}{2}\right) \cos \left(\frac{75^{\circ}-15^{\circ}}{2}\right)$

$\cos 75^{\circ}+\cos 15^{\circ}=2 \cos \left(45^{\circ}\right) \cos \left(30^{\circ}\right)$

$\cos 75^{\circ}+\cos 15^{\circ}=2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)$

$\cos 75^{\circ}+\cos 15^{\circ}=\frac{\sqrt{3}}{\sqrt{2}}$

Hence, the required answer is $\frac{\sqrt{3}}{\sqrt{2}}$.

Example 2: If $0<x, y<\pi$ and
$\cos x+\cos y-\cos (x+y)=\frac{3}{2}$

then $\sin x+\cos y$ is equal to.

Solution

The given equation is
$\cos x+\cos y-\cos (x+y)=\frac{3}{2}$

Observe that
$\frac{1}{2}+\frac{1}{2}-\left(-\frac{1}{2}\right)=\frac{3}{2}$

This is satisfied for
$x=y=60^{\circ}$

So,
$\sin 60^{\circ}+\cos 60^{\circ}=\frac{\sqrt{3}}{2}+\frac{1}{2}$

OR

$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)-\left[2 \cos ^2\left(\frac{x+y}{2}\right)-1\right]=\frac{3}{2}$

$2 \cos \left(\frac{x+y}{2}\right)\left[\cos \left(\frac{x-y}{2}\right)-\cos \left(\frac{x+y}{2}\right)\right]=\frac{1}{2}$

$2 \cos \left(\frac{x+y}{2}\right)\left[2 \sin \left(\frac{x}{2}\right)\sin \left(\frac{y}{2}\right)\right]=\frac{1}{2}$

$\cos \left(\frac{x+y}{2}\right)\sin \left(\frac{x}{2}\right)\sin \left(\frac{y}{2}\right)=\frac{1}{8}$

$x=y=60^{\circ}$

Hence, the required answer is
$\frac{1+\sqrt{3}}{2}$.

Example 3: The value of
$\cos (\alpha+\beta+\gamma)+\cos (\gamma+\alpha-\beta)+\cos (\alpha+\beta-\gamma)+\cos (\beta+\gamma-\alpha)$ is.

Solution

$\cos (\alpha+\beta+\gamma)+\cos (\alpha+\beta-\gamma)+\cos (\gamma+\alpha-\beta)+\cos (\beta+\gamma-\alpha)$

$=2 \cos \gamma \cos (\alpha+\beta)+2 \cos \gamma \cos (\alpha-\beta)$

$=2 \cos \gamma \cdot 2 \cos \alpha \cos \beta$

$=4 \cos \alpha \cos \beta \cos \gamma$

Hence, the required answer is
$4 \cos \alpha \cos \beta \cos \gamma$.

Example 4: The value of $\sin 75^{\circ}-\sin 15^{\circ}$ is.

Solution

Using the trigonometric formula,
$\sin x-\sin y=2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)$

$\sin 75^{\circ}-\sin 15^{\circ}=2 \cos \left(\frac{75^{\circ}+15^{\circ}}{2}\right) \sin \left(\frac{75^{\circ}-15^{\circ}}{2}\right)$

$\sin 75^{\circ}-\sin 15^{\circ}=2 \cos \left(45^{\circ}\right) \sin \left(30^{\circ}\right)$

$\sin 75^{\circ}-\sin 15^{\circ}=2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)$

$\sin 75^{\circ}-\sin 15^{\circ}=\frac{1}{\sqrt{2}}$

Hence, the required answer is $\frac{1}{\sqrt{2}}$.

Example 5: Find the value of
$\sin 55^{\circ}+\sin 65^{\circ}+\sqrt{3} \cos 175^{\circ}$.

Solution

$\sin 55^{\circ}+\sin 65^{\circ}+\sqrt{3} \cos 175^{\circ}$

$=2 \sin \left(\frac{55^{\circ}+65^{\circ}}{2}\right) \cos \left(\frac{55^{\circ}-65^{\circ}}{2}\right)+\sqrt{3} \cos 175^{\circ}$

$=2 \sin 60^{\circ} \cos (-5^{\circ})+\sqrt{3} \cos (180^{\circ}-5^{\circ})$

$=\sqrt{3} \cos (-5^{\circ})-\sqrt{3} \cos 5^{\circ}$

$=0$

Hence, the required answer is 0.

List of Topics Related to the Trigonometric Functions

This section lists the important topics related to Trigonometric Functions, helping you connect core concepts like ratios, identities, and angle transformations for quick revision and exam-focused preparation.

NCERT Resources

This section brings together essential NCERT Class 11 Chapter 3 resources to help you build strong fundamentals in Trigonometric Functions. It includes concise notes, detailed solutions, and exemplar problems, all aligned with the CBSE syllabus and useful for school exams as well as competitive exam preparation.

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 Sum to Product Formulae

This section includes focused practice questions on Sum to Product formulas, designed to improve your ability to convert trigonometric sums into products and apply these identities confidently in problem-solving.

Sum To Product And Product To Sum Formulas - Practice Question

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