Problem 291 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } \tan \beta=\cos \theta \tan \alpha, \text { then prove that } \sin (\alpha-\beta)=\tan ^{2} \frac{\theta}{2} \sin (\alpha+\beta) $$

Step-by-Step Solution

Answer

This exercise mainly revolves around manipulation of different trigonometric identities to get the required format of \(\sin(\alpha-\beta) = \tan^2 (\frac{\theta}{2}) \sin(\alpha +\beta)\), making a correct use of basic trigonometric formulas and identities.

Step 1: Re-write tangent using sine and cosine

Start by rewriting \(\tan \beta\) as \(\frac{\sin \beta}{\cos \beta}\). This gives us \(\frac{\sin \beta}{\cos \beta} = \cos \theta \tan \alpha\).

Step 2: Re-write right-hand side of equation

Re-write the right-hand side of the equation using the double angle formula: \(\sin 2x = 2 \sin x \cos x\). We have: \(\tan ^{2} \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\).

Step 3: Isolate the terms involving alpha and theta

Now isolate the \(\sin \alpha\) and \(\cos \alpha\) terms. This can be done by multiplying through by \(\sin \beta\): \[\sin \alpha = \frac{\sin \beta}{\cos \theta} \cos \alpha \] and then rearranging to get: \[\sin \alpha - \sin \beta \cos \theta = \cos \alpha \cos \theta\]

Step 4: Transform into required format

Use the identity \(\sin(a - b) = \sin a \cos b - \cos a \sin b\) hopefully leading to the form \(\sin (\alpha-\beta)=\tan ^{2} \frac{\theta}{2} \sin (\alpha+\beta)\).

Key Concepts

Tangent to Sine and Cosine Conversion

When delving into the realm of trigonometry, understanding how to convert from one trigonometric function to another is pivotal. A common transformation is converting tangent to its sine and cosine components. The tangent of an angle, commonly represented as \( \tan \theta \), is defined as the ratio of the sine of the angle to the cosine of the angle: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \].

This equality is incredibly useful when we need to simplify an expression involving tangents by breaking them into more fundamental trigonometric functions that are often easier to manipulate. For instance, if an equation includes \( \tan \beta \) and we wish to integrate sine and cosine into the equation, we may express \( \tan \beta \) as \( \frac{\sin \beta}{\cos \beta} \). This step is the foundational block of solving many trigonometric identities and is crucial in proving more complex relationships between trigonometric functions.

Double Angle Formulas

The double angle formulas are a cornerstone in trigonometry, providing a way to express trigonometric functions of double angles, like \( 2\theta \), in terms of single angles. These formulas stem from the sum of angles formulas and can be used to simplify and solve a variety of trigonometric problems.

The double angle formula for sine, for example, states that \( \sin(2x) = 2\sin(x)\cos(x) \). Similarly, for cosine, there are a few variations, including \( \cos(2x) = \cos^2(x) - \sin^2(x) \), \( \cos(2x) = 2\cos^2(x) - 1 \), and \( \cos(2x) = 1 - 2\sin^2(x)\). These formulas often appear in trigonometric identity proofs and can be indispensable tools for both transforming and simplifying equations, as they connect the trigonometric functions of an angle with those of its double.

Trigonometric Identity Transformations

Transforming trigonometric identities involves manipulating and rewriting trigonometric expressions using well-established identities. This is crucial when proving a given trigonometric equation like \( \sin(\alpha - \beta) \) in terms of other trigonometric functions. A powerful technique in these transformations is using sum and difference formulas, such as \( \sin(a - b) = \sin a \cos b - \cos a \sin b \) and \( \cos(a + b) = \cos a \cos b - \sin a \sin b \).

These formulas allow us to express trigonometric functions of combined angles in terms of the functions of individual angles, which are often easier to work with. The ability to shift between different forms of trigonometric expression is the essence of solving many trigonometry problems. It enables a deeper understanding of how these functions interrelate and provides a methodical approach to tackling complex trigonometric equations.