Introduction

To clearly understand why tan(1) is greater than arctan(1), it is essential to first comprehend the definitions and properties of these trigonometric functions. This article will delve into the mathematical reasoning behind this inequality, using calculus and graphical analysis to provide a comprehensive explanation.

Understanding Tan1 and Arctan1

1. Tangent Function (Tan1)

Tan1 represents the tangent of 1 radian. The tangent function is defined as the ratio of the opposite side to the adjacent side in a right triangle. It is a periodic function with a period of π, and its value increases as its argument increases.

2. Arctangent Function (Arctan1)

Arctan1, also known as the inverse tangent function, returns the angle whose tangent is 1. By definition, the value of arctan(1) is π/4 radians or 45 degrees, as the tangent of π/4 is 1.

Comparing Tan1 and Arctan1

To compare tan(1) and arctan(1), let's calculate their numerical values:

tan(1) ≈ 1.5574 (using a calculator or reference table) arctan(1) π/4 ≈ 0.7854

Thus, we can see that tan(1) is greater than arctan(1), as:

1.5574 0.7854

Extending the Concept to Arctan1

To further illustrate the relationship between the trigonometric functions, let's consider the inverse sine function (arcsine) and its comparison with the tangent function.

1. Inverse Sine Function (Arcsin1)

The arcsine function, denoted as arcsin(1), is the inverse of the sine function. Since the sine of π/2 (or 90 degrees) equals 1, we have:

arcsin(1) π/2 radians or 90 degrees

2. Tangent of 45 Degrees (Tan45°)

The tangent of 45 degrees equals 1, i.e., tan(45°) 1. The inverse of this tangent function, denoted as arctan(1), gives us 45 degrees or π/4 radians:

arctan(1) 45° π/4 radians

3. Tangent of 1 (Tan1)

The tangent of 1 radian is approximately 0.01745. Therefore, arcsin(1) is greater than tan(1) because:

arcsin(1) π/2 ≈ 1.5708

And, tan(1) approx; 0.01745

Thus, arcsin(1) tan(1)

Mathematical Analysis

To provide a more rigorous understanding, we will now use calculus to analyze the relationship between tan(x) and arctan(x) in the interval 0 ≤ x ≤ π/2.

1. Calculus-Based Approach

Consider the function f(x) tan(x) - arctan(x) in the interval 0 ≤ x ≤ π/2. Both tan(x) and arctan(x) are continuous and differentiable in this interval.

Let's find the derivative of f(x) to determine its monotonicity:

f'(x) sec^2(x) - 1/(1 x^2)

Since sec^2(x) ≥ 1 and 1/(1 x^2) ≤ 1 for 0 ≤ x ≤ π/2, it follows that f'(x) ≥ 0 in this interval. Therefore, f(x) is an increasing function.

Evaluating f(0) gives us:

f(0) tan(0) - arctan(0) 0

Since f(x) is increasing and f(0) 0, it follows that f(x) 0 for x 0. Hence, tan(x) arctan(x) in the interval 0 ≤ x ≤ π/2.

2. Graphical Approach

A graphical analysis further reinforces this result. The graph of y tan(x) lies above the line y x in the interval 0 ≤ x ≤ π/2. Since y arctan(x) is the reflection of y tan(x) in the line y x, it follows that y arctan(x) lies below the line y x in the same interval. Therefore, tan(x) arctan(x) in the interval 0 ≤ x ≤ π/2.