Problem 211 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \left.\sin x>-\frac{1}{2} \text { \\{Ans. }\left(2 n \pi-\frac{\pi}{6}, 2 n \pi+\frac{7 \pi}{6}\right)\right\\} $$

Step-by-Step Solution

Answer

The solution set of the inequality \(\sin x > -\frac{1}{2}\) is \(\left(2 n \pi - \frac{\pi}{6}, 2 n \pi + \frac{7\pi}{6}\right)\), where \(n\) is any integer.

Step 1: Understand the \(\sin x\) function

The \(\sin x\) function is a periodic function that oscillates between -1 and 1. When \(x = -\frac{\pi}{6}\), \(\sin x\) equals -1/2. Conversely, the \(\sin x\) value at \(x = \frac{7\pi}{6}\) is also -1/2.

Step 2: Solve the inequality in the main interval

Consider the interval \([0, 2\pi]\). In this interval, the inequality \(\sin x > -\frac{1}{2}\) is true when \(x\) is in the intervals \((0, \frac{7\pi}{6}]\) and \([\frac{11\pi}{6}, 2\pi)\).

Step 3: Extend the solution for all values of x

Since \(\sin x\) is a periodic function with a period of \(2\pi\), the solution is valid for all similar intervals in the form of \(\left(2 n \pi - \frac{\pi}{6}, 2 n \pi + \frac{7\pi}{6}\right)\), where \(n\) is any integer.

Key Concepts

Sine Function

The sine function, denoted as \( \sin x \), is a crucial component of trigonometry. It relates to the ratio of the opposite side to the hypotenuse in a right-angled triangle. This function oscillates between -1 and 1 as \( x \) varies.

• At \( x = 0 \), \( \sin x = 0 \).
• At \( x = \pi/2 \), \( \sin x = 1 \).
• At \( x = \pi \), \( \sin x = 0 \).
• At \( x = 3\pi/2 \), \( \sin x = -1 \).
• And finally, at \( x = 2\pi \), \( \sin x \) returns to 0.

Understanding these points helps you graph the sine wave, showing its peaks, troughs, and zeros.

Periodic Function

Periodic functions repeat their values over regular intervals, known as periods. The sine function is periodic with a common period of \( 2\pi \).

• This means that every \( 2\pi \) units along the x-axis, the sine function begins a new cycle.
• Key intervals, like from 0 to \( 2\pi \), allow us to predict behavior beyond just one cycle.
• The formula \( \sin(x + 2\pi) = \sin x \) represents this repeating nature.

When an inequality involves a periodic function, solutions must be applied repeatedly at each cycle, as seen with this exercise.

Trigonometric Inequality

A trigonometric inequality involves expressions like \( \sin x > -\frac{1}{2} \). Analyzing such inequalities means identifying intervals where the inequality holds true.

• On [0, 2\pi], find values where \( \sin x \) is greater than a specific value, here \(-\frac{1}{2}\).
• Solutions in this interval are useful for generalizing to other cycles.
• This inequality expands across all cycles by using the periodicity of \( \sin x \).

It’s always key to look for where sin crosses target values within one period and apply it across others.

Interval Notation

Interval notation is a concise way of expressing a set of numbers, typically used to describe solutions.

• The solution from the exercise, \((2n\pi - \frac{\pi}{6}, 2n\pi + \frac{7\pi}{6})\), uses intervals where sin remains greater than \(-\frac{1}{2}\).
• Parentheses \(( \; )\) denote exclusion of endpoints, while brackets \([ \; ]\) include them.
• The notation helps quickly represent infinite solutions by modifying \( n \), an integer that reflects repetitions.

Mastering interval notation simplifies solutions and enhances clarity, especially when dealing with inequalities in periodic contexts.