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.
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.
