Problem 32 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve the following inequalities: $$ \sin ^{-1} x>\cos ^{-1} x $$

Step-by-Step Solution

Answer

The solution to the inequality \(\sin ^{-1} x>\cos ^{-1} x\) is \(x > \frac{\sqrt{2}}{2}\). It must be noted that this solution is valid only for \(x\) within the range of the inverse sine and cosine function, that is, \(-1 \leq x \leq 1\).

Step 1: Interpret the inequality

Rewrite the inequality as \(\sin^{-1}(x) - \cos^{-1}(x) > 0\). This simplification will help draw comparison between the two functions.

Step 2: Simplification using trigonometric identities

Knowing the relationship between the two inverse trigonometric functions, replace \(\cos^{-1}(x)\) with \(\frac{\pi}{2} - \(\sin^{-1}(x)\). The inequality then simplifies to \(\sin^{-1}(x) - (\(\frac{\pi}{2} - \(\sin^{-1}(x)\)) > 0\). This further simplifies to \(2\sin^{-1}(x) - \frac{\pi}{2} > 0\).

Step 3: Solve for x

Rearrange the inequality to solve for \(x\). The inequality becomes \(2\sin^{-1}(x) > \frac{\pi}{2}\). Divide by 2 on both sides, landing with \(\sin^{-1}(x) > \frac{\pi}{4}\). Finally, apply the \(\sin\) function to both sides to get \(x > \sin(\frac{\pi}{4})\). Since \(\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\), therefore, \(x > \frac{\sqrt{2}}{2}\).

Key Concepts

Inequalities

When solving inequalities, the goal is to determine the set of values for the variable that makes the inequality true. In other words, instead of finding just one solution, we often seek an entire range of values. Inequalities can involve a variety of mathematical operations, including addition, subtraction, and division, and may also incorporate functions such as linear and quadratic functions. They are denoted using symbols such as `>` for greater than, `<` for less than, `≥` for greater than or equal to, and `≤` for less than or equal to.

To effectively solve an inequality:

• Simplify both sides as much as possible.
• Apply algebraic manipulation to isolate the variable of interest on one side of the inequality.
• Consider the domain of the functions involved, as certain functions might have restrictions.
• Test potential solutions and boundary values to verify the inequality.

Understanding the basic principles of inequalities helps deal with more complex functions like the inverse trigonometric functions, seen in the presented problem.

Arcsin Function

The arcsin function, represented as \(\sin^{-1}(x)\), plays a crucial role in trigonometry. It is the inverse function of the sine function. The primary purpose of the arcsin function is to give an angle whose sine is a given number. For example, if \(\sin(\theta) = x\), then \(\theta = \sin^{-1}(x)\).

Some key characteristics include:

• The domain is limited to \(-1 \leq x \leq 1\) because the sine of an angle cannot exceed these values.
• The range of arcsin is \(-\frac{\pi}{2} \leq \sin^{-1}(x) \leq \frac{\pi}{2}\).
• Arcsin is commonly used to find angles in right-angled triangles where the opposite side and hypotenuse are known.

In solving the given inequality, the arcsin function was central to expressing one side of the inequality, simplifying the process.

Arccos Function

The arccos function, denoted as \(\cos^{-1}(x)\), is the inverse of the cosine function. It provides an angle whose cosine equals a specified value. This is particularly useful in trigonometry when you know the adjacent side and hypotenuse of a right-angled triangle and need to determine the angle.

Important characteristics of the arccos function:

• The domain also ranges from \(-1 \leq x \leq 1\) for similar reasons as the arcsin function.
• Its range is \(0 \leq \cos^{-1}(x) \leq \pi\).
• It is especially helpful when analyzing angles in the context of circles or periodic functions.

In the problem, understanding how to transform \(\cos^{-1}(x)\) using trigonometric identities helped simplify the inequality, eventually leading to the solution of finding values of \(x\) where the inequality holds true.