Problem 53 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve: \(\theta=\tan ^{-1}\left(2 \tan ^{2} \theta\right)-\frac{1}{2} \sin ^{-1}\left(\frac{3 \sin 2 \theta}{5+4 \cos 2 \theta}\right)\).

Step-by-Step Solution

Answer

The steps above will help you in solving the equation. Once you will substitute the values for \(A\) and \(B\), you will get an expression for \(\theta\). Remember to check if the solution lies in the domain of the given inverse functions.

Step 1: Simplify the Expression

Initially, we will simplify the complicated expression to a more familiar form. We will need to replace the double-angle functions \(\sin 2\theta\) and \(\cos 2\theta\) with their identities. Using the fact that \(\sin 2\theta = 2\sin\theta\cos\theta\) and \(\cos 2\theta = 2\cos^2\theta - 1\) or \(\cos 2\theta = 1 - 2\sin^2\theta\), we can rewrite the equation into something we could work better with.

Step 2: Getting Rid of the Inverse Trigonometric Functions

Before we can proceed, we need to eliminate the inverse trigonometric functions in the equation. For \(\tan^{-1}\left(2 \tan^{2} \theta\right)\), we will let its argument, \(2 \tan^{2} \theta\), be some new variable \(A\), such that \(2 \tan^{2} \theta = A\) and hence \(\tan^{-1}\left(2 \tan^{2} \theta\right) = A\). Similarly for the second part of the equation, let \(B = \frac{3 \sin 2 \theta}{5+4 \cos 2 \theta}\) so that \(\sin^{-1}\left(\frac{3 \sin 2 \theta}{5+4 \cos 2 \theta}\right) = B\), and rewrite the equation again.

Step 3: Solving the Equation

After simplifying the equation, we now have \(A - \frac{1}{2}B = \theta\). Substitute the values of \(A\) and \(B\) from step 2 into the equation and proceed to solve for \(\theta\). Here you might need to use one of the trigonometric identities or some algebraic manipulation to find the solution.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the functions used to obtain an angle from a trigonometric ratio. The most common inverse trigonometric functions include

• \( \sin^{-1}(x) \)
• \( \cos^{-1}(x) \)
• \( \tan^{-1}(x) \)

These are often termed as arcsin, arccos, and arctan, respectively. They are critical when you need to solve equations where the angle is the unknown. In the given problem, we have two inverse trig functions, \( \tan^{-1}(2 \tan^{2} \theta) \) and \( \sin^{-1}\left(\frac{3 \sin 2 \theta}{5+4 \cos 2 \theta}\right) \). These functions allow us to express angles in terms of known values or other functions, like sine, cosine, and tangent.
An important aspect of using inverse trig functions is ensuring the inputs (the range of the trigonometric ratios) satisfy the domains of these functions, which in turn ensures the outputs (angles) remain valid and within acceptable limits.

Double-Angle Identities

Double-angle identities are specific trigonometric identities that provide alternate expressions for the sine, cosine, and tangent of doubled angles. These identities are particularly useful in simplifying trigonometric expressions and solving trigonometric equations. They are:

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

In our exercise, these identities help break down complicated terms into simpler ones, paving the way for easier algebraic manipulation. By replacing \( \sin 2\theta \) and \( \cos 2\theta \) with their identity forms, the given equation can be greatly simplified, allowing for a more straightforward solving process. It's crucial to choose the right identity based on the available expression for effective simplification.

Algebraic Manipulation

Algebraic manipulation involves the rearrangement and simplification of equations or expressions using various algebraic rules and techniques. This process is essential when solving trigonometric equations as it helps in isolating the variable of interest. The original exercise requires such techniques to manage trigonometric functions and expressions efficiently.

• We start by identifying substitutions to simplify expressions, like letting \( A = 2\tan^2\theta \) and \( B = \frac{3 \sin 2 \theta}{5+4 \cos 2 \theta} \), as noted in the solution steps.
• By organizing and substituting these values back into the equation, solving becomes a series of simplification steps, often involving standard algebraic techniques like distribution, factorization, and rearrangement.

These manipulations ultimately lead us to a simpler equation, such as \( A - \frac{1}{2}B = \theta \), where we can solve for \( \theta \) using the isolated terms. Mastery of these techniques is crucial in handling complex mathematical problems efficiently.