Problem 21 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve: \(\sin 7 x+\sin 4 x+\sin x=0,0 \leq x \leq \frac{\pi}{2}\)

Step-by-Step Solution

Answer

The solutions to the given trigonometric equation within the specified range are \(x=\frac{\pi}{9}\) and \(x=\frac{5\pi}{9}\).

Step 1: Apply Trigonometric Identities

Express the equation using the sum-to-product identities. These identities are \(\sin a + \sin b = 2 \sin \frac{1}{2}(a+b) \cos \frac{1}{2}(a-b)\) and \(\sin a - \sin b = 2 \cos \frac{1}{2}(a+b) \sin \frac{1}{2}(a-b)\). Apply these to the given problem to get, \(\sin 7x + \sin x = \sin 4x\). Applying sum-to-product identity, this simplifies to \(2 \sin 4x \cos 3x = \sin 4x\).

Step 2: Simplify equation

The equation simplifies further by dividing both sides by \(\sin 4x\), we get \(2 \cos 3x = 1\).

Step 3: Solve for x within the given range

Now solve \(2 \cos 3x = 1\), we have \(\cos 3x = \frac{1}{2}\). Since \(0 \leq x \leq \frac{\pi}{2}\), \(0 \leq 3x \leq \frac{3\pi}{2}\). Now, Cosine function gives \(\frac{1}{2}\) at \(x=\frac{\pi}{3}\) and \(x=\frac{5\pi}{3}\) in the interval \([0, 2\pi]\). Therefore, the solutions are \(x=\frac{\pi}{9}\) and \(x=\frac{5\pi}{9}\).

Key Concepts

Sum-to-Product Identities

In trigonometry, sum-to-product identities are crucial for transforming the sum or difference of sines or cosines into a product, simplifying complex equations. This can make handling and solving trigonometric equations much easier. Here are the two key identities:

• For sine: \( \sin a + \sin b = 2 \sin \left( \frac{1}{2}(a+b) \right) \cos \left( \frac{1}{2}(a-b) \right) \)
• For sine differences: \( \sin a - \sin b = 2 \cos \left( \frac{1}{2}(a+b) \right) \sin \left( \frac{1}{2}(a-b) \right) \)

In our problem, the identity helps convert the sum of sine terms into products, ultimately reducing complexity. The original equation \(\sin 7x + \sin 4x + \sin x = 0\) is first broken down using this identity, allowing the problem to be tackled more effectively.

Trigonometric Equations

Trigonometric equations involve trigonometric functions, which are solved for specific angles within a given interval. These equations are common in calculus and physics due to their periodic nature. When solving such equations, it's important to:

• Simplify the equation using known identities.
• Find general solutions before narrowing to specific intervals.

In the provided exercise, the equation \(2 \sin 4x \cos 3x = \sin 4x \) needed to be simplified. By isolating and simplifying the components, we helped pave the path towards finding the solutions efficiently. Trigonometric equations often require multiple steps to simplify before arriving at a solution.

Cosine Function

The cosine function, denoted as \( \cos \), is a periodic function with a cycle of \(2\pi\), possessing values from -1 to 1. In this exercise, understanding the behavior of the cosine function is crucial, as it assists in solving the equation \(2 \cos 3x = 1\). The aim was to reach the point where we can easily solve for \(x\).

• Cosine equals \( \frac{1}{2} \) at angles \(x=\frac{\pi}{3}\) and \(x=\frac{5\pi}{3}\) within one cycle \([0, 2\pi]\).
• These roots help in determining specific solutions within given bounds.

Mastering cosine function properties aids in pinpointing solutions over a specified range, and is instrumental when dealing with period overlaps and transformations.

Trigonometric Solutions

Finding solutions to trigonometric problems involves identifying angles where a function reaches specific values. After simplifying the problem with identities, the next goal is to solve equations like \(\cos 3x = \frac{1}{2}\). Success requires:

• Recognizing key values where trigonometric functions achieve given outputs.
• Utilizing properties and periodic behaviors effectively.

For this equation, potential solutions within an interval \([0, \frac{3\pi}{2}]\) were derived from known cosine roots. By focusing on these trigonometric solution techniques, students can become adept at solving these types of equations.

Interval Analysis

Given a set boundary for solutions, interval analysis simplifies the identification of valid answers within a range. For example, finding solutions within interval \([0, \frac{\pi}{2}]\) requires extra care to ensure all solutions meet interval restrictions.

• Verify each potential solution fits within the specific bounds.
• Consider the effect of trigonometric function periodicity on intervals.

In our exercise, calculated cosine roots belonged to a larger span, \([0, \frac{3\pi}{2}]\), but were refined to meet the \([0, \frac{\pi}{2}]\) interval. This analysis ensures only accurate, relevant solutions are considered. Understanding how to confine solutions within given intervals is an essential skill in mathematical problem-solving.