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    Product To Sum Formulas

    Product To Sum Formulas

    Hitesh SahuUpdated on 10 Feb 2026, 05:36 PM IST

    Think about multiplying two sounds or waves together - what you get is often easier to understand when you break it into simpler parts. In trigonometry, a similar idea applies: product-to-sum formulas help us rewrite the product of trigonometric functions as a sum or difference, making calculations much simpler. In this article on Product to Sum Formulas, we will learn what these formulas are, why they are useful, and how to apply them to simplify trigonometric expressions and solve problems, especially for Class 11 mathematics, board exams, and competitive exams like JEE.

    This Story also Contains

    1. What Are Product-to-Sum and Product-to-Difference Formulas?
    2. Proof of Product-to-Sum Formulas in Trigonometry
    3. Standard Sum and Difference Identities Used
    4. Applications of Product to Sum Formula
    5. Solved Examples Based on Product to Sum/Difference
    6. List of Topics Related to the Trigonometric Functions
    7. NCERT Resources
    8. Practice Questions on Product to Sum Formulae
    Product To Sum Formulas
    Product To Sum Formulas

    What Are Product-to-Sum and Product-to-Difference Formulas?

    Product-to-sum formulas in trigonometry are identities used to convert the product of sine and cosine functions into a sum or difference of trigonometric functions. These formulas are derived by adding or subtracting the sum and difference trigonometric identities of sine and cosine.

    Product-to-sum identities are extremely useful for simplifying trigonometric expressions, solving equations involving products of trig functions, and handling integration and series problems in higher mathematics.

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

    The product-to-sum formulas allow us to rewrite products of trigonometric ratios as sums or differences.
    Let $\alpha$ and $\beta$ be any two angles.

    1. Product of Cosine Functions

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

    This identity converts the product of two cosine functions into the sum of cosine functions.

    2. Product of Sine Functions

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

    This formula expresses the product of two sine functions as the difference of cosine functions.

    3. Product of Sine and Cosine Functions

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

    This identity transforms the product of sine and cosine into the sum of sine functions.

    4. Product of Cosine and Sine Functions

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

    This formula converts the product of cosine and sine into the difference of sine functions.

    Proof of Product-to-Sum Formulas in Trigonometry

    The product-to-sum identities are derived using the sum and difference formulas of sine and cosine. These proofs are important for understanding how products of trigonometric functions can be converted into sums or differences, a key skill for Class 11 trigonometry, board exams, and competitive exams like JEE.

    Standard Sum and Difference Identities Used

    We begin with the fundamental trigonometric identities:

    $\sin(A+B)=\sin A\cos B+\cos A\sin B$  (1)

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

    $\cos(A+B)=\cos A\cos B-\sin A\sin B$  (3)

    $\cos(A-B)=\cos A\cos B+\sin A\sin B$  (4)

    1. Proof of $\sin A\cos B=\frac{1}{2}[\sin(A+B)+\sin(A-B)]$

    Adding equations (1) and (2):

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

    Dividing both sides by $2$:

    $\sin A\cos B=\frac{1}{2}[\sin(A+B)+\sin(A-B)]$

    2. Proof of $\cos A\sin B=\frac{1}{2}[\sin(A+B)-\sin(A-B)]$

    Subtracting equation (2) from equation (1):

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

    Dividing both sides by $2$:

    $\cos A\sin B=\frac{1}{2}[\sin(A+B)-\sin(A-B)]$

    3. Proof of $\cos A\cos B=\frac{1}{2}[\cos(A+B)+\cos(A-B)]$

    Adding equations (3) and (4):

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

    Dividing both sides by $2$:

    $\cos A\cos B=\frac{1}{2}[\cos(A+B)+\cos(A-B)]$

    4. Proof of $\sin A\sin B=\frac{1}{2}[\cos(A-B)-\cos(A+B)]$

    Subtracting equation (3) from equation (4):

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

    Dividing both sides by $2$:

    $\sin A\sin B=\frac{1}{2}[\cos(A-B)-\cos(A+B)]$

    Applications of Product to Sum Formula

    • Used to simplify trigonometric expressions involving products of sine and cosine functions.

    • Helps in evaluating trigonometric values of non-standard angles like $15^\circ$, $18^\circ$, or $75^\circ$.

    • Frequently applied in solving trigonometric equations by converting products into simpler sums or differences.

    • Useful in proving trigonometric identities by reducing complex products into manageable forms.

    • Plays an important role in Class 11–12 board exams and competitive exams such as JEE and NDA.

    • Applied in physics and engineering problems involving wave motion, oscillations, and signal analysis.

    Solved Examples Based on Product to Sum/Difference

    Example 1: The value of $\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{6\pi}{7}\right)$ is equal to? [JEE MAINS 2022]

    Solution:

    Using the summation of the cosine series,

    $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}$

    $=\dfrac{\sin\left(\frac{3\times2\pi}{2\times7}\right)}{\sin\left(\frac{2\pi}{2\times7}\right)}\times\cos\left(\frac{2\pi/7+6\pi/7}{2}\right)$

    $=\dfrac{\sin\frac{5\pi}{7}}{\sin\frac{\pi}{7}}\times\cos\frac{4\pi}{7}$

    $=\dfrac{\sin\left(\pi-\frac{4\pi}{7}\right)\cos\frac{4\pi}{7}}{\sin\frac{\pi}{7}}$

    $=\dfrac{2\sin\frac{4\pi}{7}\cos\frac{4\pi}{7}}{2\sin\frac{\pi}{7}}$

    $=\dfrac{\sin\frac{8\pi}{7}}{2\sin\frac{\pi}{7}}$

    $=-\dfrac{1}{2}$

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

    Example 2: The value of $\cos^2 10^\circ-\cos10^\circ\cos50^\circ+\cos^2 50^\circ$ is [JEE MAINS 2019]

    Solution:

    $\cos^2 10^\circ-\cos10^\circ\cos50^\circ+\cos^2 50^\circ$

    $=\dfrac{1+\cos20^\circ}{2}+\dfrac{1+\cos100^\circ}{2}-\cos10^\circ\cos50^\circ$

    $=\dfrac{1}{2}[2+\cos20^\circ+\cos100^\circ-2\cos10^\circ\cos50^\circ]$

    $=\dfrac{1}{2}[2+\cos100^\circ+\cos20^\circ-\cos60^\circ-\cos40^\circ]$

    $=\dfrac{1}{2}\left[\dfrac{3}{2}+2\cos60^\circ\cos40^\circ-\cos40^\circ\right]$

    $=\dfrac{3}{4}$

    Hence, the answer is $\frac{3}{4}$.

    Example 3: If $x+\frac{1}{x}=2\cos\theta$, then find $x^3+\frac{1}{x^3}$.

    Solution:

    $(x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3x\cdot\frac{1}{x}(x+\frac{1}{x})$

    $x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3(x+\frac{1}{x})$

    Given $x+\frac{1}{x}=2\cos\theta$,

    $x^3+\frac{1}{x^3}=(2\cos\theta)^3-3(2\cos\theta)$

    $=2(4\cos^3\theta-3\cos\theta)$

    $=2\cos3\theta$

    Hence, the answer is $2\cos3\theta$.

    Example 4: The value of $\sin18^\circ+\sin72^\circ-2\cos27^\circ$ is

    Solution:

    $\sin18^\circ+\sin72^\circ-2\cos27^\circ$

    $=2\sin45^\circ\cos27^\circ-2\cos27^\circ$

    $=0$

    Hence, the answer is 0.

    Example 5: If $\sin(3x)+\sin(2x)-\sin(x)=0$, then find the number of solutions in $[0,\pi]$.

    Solution:

    $\sin(3x)+\sin(2x)-\sin(x)=0$

    $\sin(3x)-\sin(x)+\sin(2x)=0$

    $2\sin x\cos2x+\sin2x=0$

    $2\sin x\cos2x+2\sin x\cos x=0$

    $2\sin x(\cos2x+\cos x)=0$

    $2\sin x\cdot2\cos\frac{3x}{2}\cos\frac{x}{2}=0$

    $x=0, \frac{\pi}{3},\pi$

    Hence, the answer is 3.

    List of Topics Related to the Trigonometric Functions

    This section gives a quick overview of important topics related to trigonometric functions, helping you see how different concepts are connected and making revision easier for exams and practice.

    NCERT Resources

    This section provides essential NCERT-based study resources for Class 11 Trigonometric Functions, including concise notes, step-by-step solutions, and exemplar problems to support clear understanding and exam-focused 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 Product to Sum Formulae

    This section offers targeted practice questions on Product to Sum formulas, helping you strengthen simplification skills and apply trigonometric identities effectively in exam-oriented problems.

    Product To Sum Formulas - Practice Question

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


    Frequently Asked Questions (FAQs)

    Q: What are Product to Sum formulas in trigonometry?
    A:

    Product to Sum formulas convert the product of sine and cosine functions into the sum or difference of trigonometric functions, making expressions easier to simplify.

    Q: Which trigonometric products can be converted using Product to Sum formulas?
    A:

    These formulas apply to:

    $\sin A \sin B$

    $\cos A \cos B$

    $\sin A \cos B$

    $\cos A \sin B$

    Q: Are Product to Sum formulas derived from basic identities?
    A:

    Yes, they are derived using the sum and difference identities of sine and cosine.

    Q: Can Product to Sum formulas be used to evaluate numerical values?
    A:

    Yes, they are especially useful for evaluating expressions involving non-standard angles like $15^\circ$, $18^\circ$, or $75^\circ$.

    Q: Do Product to Sum formulas work for all angles?
    A:

    Yes, they are valid for all real angles, provided the trigonometric functions are defined.

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