Problem 24 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

\(\sin ^{-1} x>\cos ^{-1} x\) holds for (a) all values of \(x\) (b) \(x \varepsilon\left(0, \frac{1}{\sqrt{2}}\right)\) (c) \(\left(\frac{1}{\sqrt{2}}, 1\right)\) (d) \(x=0.75\)

Step-by-Step Solution

Answer

Only option (b) \(x \in (0, \frac{1}{\sqrt{2}})\) is correct.

Step 1: Understand inverse sine and cosine functions properties

The sine inverse function, \( \sin^{-1}x \), is defined for \( -1 \leq x \leq 1 \) and its range is \( -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2} \). The cosine inverse function, \( \cos^{-1}x \), is also defined for \( -1 \leq x \leq 1 \) but its range is \( 0 \leq \cos^{-1}x \leq \pi \). For \( 0 \leq x \leq 1 \), \( \sin^{-1}x \) increases from 0 to \( \frac{\pi}{2} \) and \( \cos^{-1}x \) decreases from \( \frac{\pi}{2} \) to 0. This means that \( \sin^{-1}x > \cos^{-1}x \) holds for \( x \in (0, \frac{1}{\sqrt{2}}) \).

Step 2: Verify option (a)

Option (a) suggests the inequality holds for all values of \(x\). However, as concluded in Step 1, the inequality only holds for \( x \in (0, \frac{1}{\sqrt{2}}) \). Therefore, option (a) is incorrect.

Step 3: Verify option (b)

Option (b) suggests the inequality holds for \( x \in (0, \frac{1}{\sqrt{2}}) \). As concluded in Step 1, this precisely the interval for which the inequality holds. Therefore, option (b) is correct.

Step 4: Verify option (c)

Option (c) suggests the inequality holds for \( x \in (\frac{1}{\sqrt{2}}, 1) \). However, in this interval \( \cos^{-1}x > \sin^{-1}x \), contradicting the inequality given in the problem. Therefore, option (c) is incorrect.

Step 5: Verify option (d)

Option (d) suggests the inequality holds for \( x = 0.75 \). However, \( 0.75 \) is not in the interval \( (0, \frac{1}{\sqrt{2}}) \), so the inequality does not hold. Therefore, option (d) is incorrect.

Key Concepts

Sine Inverse Function

The sine inverse function, often written as \( \sin^{-1}x \) or arcsin, helps us find the angle whose sine is a given number. This function is defined for values of \( x \) ranging from \(-1\) to \(1\). The result, or output, of the sine inverse function is measured in radians, falling within the interval of \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

The sine inverse function is significant in trigonometry because it allows us to determine angles in right triangles or solve trigonometric equations. When dealing with angles in the first quadrant, \( \sin^{-1}x \) will give positive angles increasing from 0 up to \( \frac{\pi}{2} \) as \( x \) increases from 0 to 1.

• Common angles and their sine values include \( \sin^{-1}(0) = 0 \)
• \( \sin^{-1}(0.5) = \frac{\pi}{6} \)
• \( \sin^{-1}(1) = \frac{\pi}{2} \)

Understanding how the sine inverse function works helps in solving equations involving sine and becomes essential when exploring inequalities involving inverse trigonometric functions.

Cosine Inverse Function

The cosine inverse function, denoted as \( \cos^{-1}x \) or arccos, is similar in purpose to the sine inverse function. It helps us find the angle whose cosine is a specific value. It functions over the domain of \(-1 \leq x \leq 1\) and has a range from \(0\) to \(\pi\) radians.

This function is useful for solving geometric problems involving triangles, especially when the lengths of the sides are known, but the angles need to be determined. As \( x \) increases from 0 to 1, the angle measured by \( \cos^{-1}x \) ranges from \( \frac{\pi}{2} \) down to 0. When exploring inverse trigonometric inequalities involving cosine, we need to keep track of these properties:

• \( \cos^{-1}(1) = 0 \)
• \( \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} \)
• \( \cos^{-1}(0) = \frac{\pi}{2} \)

An understanding of how to use the cosine inverse function is critical for situations where we are asked to compare or relate inverse sine and cosine outputs.

Trigonometric Inequalities

Trigonometric inequalities involve comparisons between functions like sine, cosine, and their inverses. In the given exercise, we delve into an inequality involving \( \sin^{-1}x \) and \( \cos^{-1}x \). Such inequalities often arise in problems requiring angle comparisons.

The inequality \( \sin^{-1} x > \cos^{-1} x \) holds true in certain intervals, particularly when \( x \) lies between \(0\) and \(\frac{1}{\sqrt{2}}\). This is because in this range, \( \sin^{-1}x \) increases towards \(\frac{\pi}{2}\) while \( \cos^{-1}x \) decreases towards 0, allowing for \( \sin^{-1} x \) to become greater than \( \cos^{-1} x \).

When solving trigonometric inequalities like this:

• Identify ranges for inverse trigonometric functions.
• Consider how the functions change as \(x\) moves within the domain.
• Apply known values or relationships to determine where the inequality holds.

Grasping these concepts is essential for students to accurately solve and understand problems involving comparisons of inverse trigonometric functions.