Problem 116 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(0\cos x\) (B) \(\cos (\sin x)<\cos x\) (C) \(\cos (\sin x)>\sin (\cos x)\) (D) \(\cos (\sin x)<\sin (\cos x)\)

Step-by-Step Solution

Answer

The answer is (C): \(\cos(\sin x) > \sin(\cos x)\).

Step 1: Understand the Range of sin(x) and cos(x)

Given that \(0 < x < \frac{\pi}{2}\), we know both \(\sin x\) and \(\cos x\) will lie between 0 and 1. Specifically, \(\sin x\) increases from 0 to 1, while \(\cos x\) decreases from 1 to 0 as \(x\) goes from 0 to \(\frac{\pi}{2}\).

Step 2: Analyze \(\cos(\sin x)\) vs \(\cos x\)

Since \(\sin x < x\) for \(0 < x < \frac{\pi}{2}\), and the cosine function \(\cos x\) is decreasing in this range, we can conclude that \(\cos(\sin x) > \cos x\).

Step 3: Eliminate Choices Based on Step 2

From Step 2, we know \(\cos(\sin x) > \cos x\), allowing us to eliminate options (B) and (D).

Step 4: Analyze \(\cos(\sin x)\) vs \(\sin(\cos x)\)

Consider that \(\sin(\cos x)\) is the sine of a value between 0 and 1. Since the sine function increases in this domain, \(\sin(\cos x) < \cos(\sin x)\) because \(\cos(\sin x)\) is closer to 1 due to \(\sin x\) being less than \(x\).

Step 5: Determine the Correct Answer

Based on Steps 2 and 4, we conclude that \(\cos(\sin x) > \sin(\cos x)\). Therefore, the correct option is (C).

Key Concepts

Cosine Function

The cosine function is a fundamental concept in trigonometry, dealing with the ratio of the adjacent side to the hypotenuse in a right triangle. This function is defined on all real numbers and is periodic with a period of \(2\pi\). The cosine function decreases monotonically from 1 to 0 as \(x\) progresses from 0 to \(\frac{\pi}{2}\). This behavior is important when analyzing trigonometric inequalities.

For values of \(x\) in the range \(0 < x < \frac{\pi}{2}\), the cosine function's values decrease as \(x\) increases.

• At \(x = 0\), \(\cos x = 1\).
• As \(x\) approaches \(\frac{\pi}{2}\), \(\cos x\) approaches 0.

The property of being a decreasing function in this interval is crucial for understanding why \(\cos(\sin x)\) is greater than \(\cos x\) when \(\sin x < x\), which aligns with the problem statement and helps in solving trigonometric inequalities involving the cosine function.

Sine Function

The sine function represents the ratio of the opposite side to the hypotenuse in a right triangle.
It is also periodic, with a period of \(2\pi\), though it behaves differently within its cycle.

Within the critical interval \(0 < x < \frac{\pi}{2}\), the sine function increases from 0 to 1.

• At \(x = 0\), \(\sin x = 0\).
• As \(x\) approaches \(\frac{\pi}{2}\), \(\sin x\) increases steadily towards 1.

Understanding this monotonic increase helps predict behaviors in trigonometric inequalities.
For example, since \(\sin x < x\) in this interval, it influences how sine values fit into larger trigonometric expressions like \(\cos(\sin x)\) and explains why using \(\sin(\cos x)\) is crucial for evaluating inequalities correctly in such an interval.

Inequalities in Trigonometry

Trigonometric inequalities involve comparing expressions that contain trigonometric functions like sine and cosine.
These inequalities require a deep understanding of the functions' behavior in specific intervals.

In the context of this exercise, evaluating inequalities between trigonometric functions involves understanding how the sine and cosine functions interact within the interval \(0 < x < \frac{\pi}{2}\).
To solve these inequalities:

• Identify the nature of each function in the given interval — \(\sin\) is increasing, and \(\cos\) is decreasing.
• Analyze the composite functions \(\cos(\sin x)\) and \(\sin(\cos x)\).
  For \(\cos(\sin x)\), since \(\sin x < x\), and \(\cos x\) is decreasing, we find \(\cos(\sin x) > \cos x\).
• For \(\sin(\cos x)\), notice that it is the sine of a value between 0 and 1, making \(\sin(\cos x) < \cos(\sin x)\).

This comprehensive understanding helps in selecting the correct inequality, leading to the conclusion that \(\cos(\sin x) > \sin(\cos x)\).
Such approaches are vital for solving more complex trigonometric inequalities.