Problem 73 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \frac{\sin 2 A+\sin 2 B}{\sin 2 A-\sin 2 B}=\frac{\tan (A+B)}{\tan (A-B)} $$

Step-by-Step Solution

Answer

We can prove the given expression using the double angle and compound angle formulas as follows: 1. Convert the given expression using double-angle formulas: \(\frac{\sin A \cos A + \sin B \cos B}{\sin A \cos A - \sin B \cos B}\) 2. Rewrite using compound angle formulas: \(\frac{\sin(A + B) \cos A \cos B}{\sin(A - B) \cos A \cos B}\) 3. Simplify and express it as a ratio of tangents: \(\frac{\tan(A+B)}{\tan(A-B)}\) Thus, we have proven that \(\frac{\sin 2A + \sin 2B}{\sin 2A - \sin 2B} = \frac{\tan(A+B)}{\tan(A-B)}\).

Step 1: Sum to Product Formula

Start with the given expression: \[ \frac{\sin 2A + \sin 2B}{\sin 2A - \sin 2B} \] Using the double angle formula, we have: \[ \frac{2\sin A \cos A + 2\sin B \cos B}{2\sin A \cos A - 2\sin B \cos B} \] Now, we can factor the common factor \(2\) out from both numerator and denominator: \[ \frac{\sin A \cos A + \sin B \cos B}{\sin A \cos A - \sin B \cos B} \]

Step 2: Compound Angle Formulae

Using the compound angle formulae, we rewrite the expression as: \[ \frac{(\sin A \cos B + \cos A \sin B) \cos A \cos B}{(\sin A \cos B - \cos A \sin B) \cos A \cos B} \] which simplifies to: \[ \frac{\sin(A + B) \cos A \cos B}{\sin(A - B) \cos A \cos B} \]

Step 3: Ratio of Tangents

Now, we divide both numerator and denominator by \(\cos A \cos B\): \[ \frac{\sin(A + B)}{\sin(A - B)} \] Using the definition of tangent, we rewrite this as: \[ \frac{\tan(A+B)}{\tan(A-B)} \] This is our desired result. Therefore, we have proven that: \[ \frac{\sin 2A + \sin 2B}{\sin 2A - \sin 2B} = \frac{\tan(A+B)}{\tan(A-B)} \]

Key Concepts

Sum to Product Formula

The Sum to Product Formula is a pair of equations in trigonometry that transforms a sum or difference of trigonometric functions into a product.

For sine functions, the formula is as follows:

• \( \text{For } \theta_1 \text{ and } \theta_2: \[ \sin(\theta_1) + \sin(\theta_2) = 2 \sin\left(\frac{\theta_1 + \theta_2}{2}\right) \cos\left(\frac{\theta_1 - \theta_2}{2}\right) \] \[ \sin(\theta_1) - \sin(\theta_2) = 2 \cos\left(\frac{\theta_1 + \theta_2}{2}\right) \sin\left(\frac{\theta_1 - \theta_2}{2}\right) \] \)

These formulas are instrumental when simplifying trigonometric expressions and solving complex trigonometric equations. In the context of solving the given exercise, the formula helps to convert the sum and difference of \( \sin 2A \) and \( \sin 2B \) into a product, which later facilitates the simplification of the expression.

Compound Angle Formulae

Compound Angle Formulae are key in trigonometry for finding the sine, cosine, or tangent of the sum or difference of two angles.

The formulae are expressed as:

• \( \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B \)
• \( \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B \)
• \( \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B} \)

These formulae are extremely useful when solving trigonometric problems that involve angles that are composites of two separate angles. In this exercise, the compound angle formulae allow us to simplify the terms within the numerator and denominator, further leading to a form that closely relates to the definition of tangent for the given angles.

Double Angle Formula

The Double Angle Formula refers to the specific compound angle formulae where the two angles involved are the same. This gives us a way to express trigonometric functions of twice an angle in terms of the single angle.

The key double angle formulae for sine, cosine, and tangent are:

• \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \)
• \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2(\theta) - 1 = 1 - 2\sin^2(\theta) \)
• \( \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \)

In our solution, the double angle formula is critical as it converts \( \sin 2A \) and \( \sin 2B \) into expressions involving single angles, making them amenable to the Sum to Product and Compound Angle Formulae.

Ratio of Tangents

The Ratio of Tangents refers to the instance when we take the ratio of the tangent of two angles. In general form, the ratios can be manipulated using the identity: \( \frac{\tan A}{\tan B} = \frac{\sin A/\cos A}{\sin B/\cos B} \).

This can be simplified further using the fundamental properties of trigonometric functions. In the context of trigonometric identities proofs, such as the one given in the exercise, applying the ratio of tangents can bring us to an expression that neatly portrays a relationship between angles and their tangents.
Here's a key identity involving the tangents of the sum and difference of two angles, which can be proved using compound angle formulae:

• \( \frac{\tan(A+B)}{\tan(A-B)} = \frac{\sin(A+B)/\cos(A+B)}{\sin(A-B)/\cos(A-B)} \)

In the given exercise, after simplifying the expression using the formulae mentioned previously, we ultimately express the result as the ratio of the tangent of the sum and difference of angles A and B. This final form answers the problem statement and solidifies our understanding of the intricate interplay between different trigonometric functions and identities.