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

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

Hello JEE aspirants! In this lesson, we'll explore the fascinating world of product-to-sum and sum-to-product trigonometric formulas. These formulas are extremely useful for simplifying complex trigonometric expressions and solving tricky equations, skills that are frequently tested in JEE Main. Mastering these will not only boost your score but also deepen your understanding of trigonometry.

Why These Formulas Matter?

Imagine you're faced with an integral like $\int \sin 5x \cos 3x \, dx$. Directly integrating this product might seem daunting, but using a product-to-sum formula transforms it into a much simpler form! Or consider solving an equation like $\sin 5x + \sin 3x = 0$. Sum-to-product formulas can convert this into a product, making it easier to find the solutions. Essentially, these formulas give you powerful tools for manipulating trigonometric expressions.

Conceptual Explanation and Intuition

The core idea behind these formulas is to convert multiplication operations (products) into addition/subtraction (sums) and vice-versa within trigonometric functions. This is based on the fundamental relationships that exist between different trigonometric functions, and how angles relate to each other. Let's dive into the formulas!

1. Product-to-Sum Formulas

These formulas help to express the product of two trigonometric functions as a sum or difference of trigonometric functions.

1. $2\sin A \cos B = \sin(A+B) + \sin(A-B)$

Derivation: Recall the angle sum and difference formulas: $$\sin(A+B) = \sin A \cos B + \cos A \sin B$$ $$\sin(A-B) = \sin A \cos B - \cos A \sin B$$ Adding these two equations, we get: $$\sin(A+B) + \sin(A-B) = 2\sin A \cos B$$

Example: Simplify $2 \sin 50^\circ \cos 20^\circ$. Using the formula, we get $\sin(50^\circ + 20^\circ) + \sin(50^\circ - 20^\circ) = \sin 70^\circ + \sin 30^\circ = \sin 70^\circ + \frac{1}{2}$.

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

Derivation: Again, recall the angle sum and difference formulas: $$\cos(A+B) = \cos A \cos B - \sin A \sin B$$ $$\cos(A-B) = \cos A \cos B + \sin A \sin B$$ Adding these two equations, we obtain: $$\cos(A+B) + \cos(A-B) = 2\cos A \cos B$$

Example: Simplify $2 \cos 75^\circ \cos 15^\circ$. Using the formula, we have $\cos(75^\circ + 15^\circ) + \cos(75^\circ - 15^\circ) = \cos 90^\circ + \cos 60^\circ = 0 + \frac{1}{2} = \frac{1}{2}$.

3. $2\sin A \sin B = \cos(A-B) - \cos(A+B)$

Derivation: Using the same angle sum and difference formulas for cosine as above, subtracting the equation for $\cos(A+B)$ from $\cos(A-B)$ gives us: $$\cos(A-B) - \cos(A+B) = 2\sin A \sin B$$

Example: Simplify $2 \sin 40^\circ \sin 20^\circ$. Using the formula, we get $\cos(40^\circ - 20^\circ) - \cos(40^\circ + 20^\circ) = \cos 20^\circ - \cos 60^\circ = \cos 20^\circ - \frac{1}{2}$.

2. Sum-to-Product Formulas

These formulas are used to express the sum or difference of two trigonometric functions as a product.

4. $\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$

Derivation: Let $A = x + y$ and $B = x - y$. Then, $x = \frac{A+B}{2}$ and $y = \frac{A-B}{2}$. We know that $\sin(x+y) + \sin(x-y) = 2\sin x \cos y$. Substituting $x$ and $y$, we get the desired formula.

Example: Simplify $\sin 70^\circ + \sin 10^\circ$. Using the formula, we get $2 \sin \frac{70^\circ + 10^\circ}{2} \cos \frac{70^\circ - 10^\circ}{2} = 2 \sin 40^\circ \cos 30^\circ = 2 \sin 40^\circ \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \sin 40^\circ$.

5. $\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$

Derivation: Similar to the sine case, let $A = x + y$ and $B = x - y$. Then, $x = \frac{A+B}{2}$ and $y = \frac{A-B}{2}$. We know that $\cos(x+y) - \cos(x-y) = -2\sin x \sin y$. Substituting $x$ and $y$, we arrive at the formula.

Example: Simplify $\cos 80^\circ - \cos 20^\circ$. Using the formula, we get $-2 \sin \frac{80^\circ + 20^\circ}{2} \sin \frac{80^\circ - 20^\circ}{2} = -2 \sin 50^\circ \sin 30^\circ = -2 \sin 50^\circ \cdot \frac{1}{2} = -\sin 50^\circ$.

Applications in Simplification

These formulas are powerful tools to simplify complex expressions. Look for patterns where you can apply them. Sometimes, you might need to combine these with other trigonometric identities.

Solving Equations Using These Formulas

When solving trigonometric equations, these formulas can help transform sums into products, making it easier to isolate the variable. Remember to consider the general solutions.

JEE-Specific Tricks

Trick 1: Sometimes, JEE questions may involve nested trigonometric functions. Use these formulas iteratively to simplify them layer by layer.

Trick 2: Remember to connect these formulas with inverse trigonometric functions. Questions might involve simplifying expressions like $\sin^{-1}(\sin A + \sin B)$.

Example: Solve $\sin 3x + \sin x = 0$. Using the sum-to-product formula, we get $2\sin 2x \cos x = 0$. This implies either $\sin 2x = 0$ or $\cos x = 0$. Therefore, $2x = n\pi$ or $x = (2n+1)\frac{\pi}{2}$, where $n$ is an integer. This gives us $x = \frac{n\pi}{2}$ or $x = (2n+1)\frac{\pi}{2}$. Now, we would need to check these solutions against the original equation.