Understanding the Equality of cos(3nπ/2) and cos(nπ/2): A Detailed Analysis

Introduction

Trigonometric functions, such as cosine, are fundamental in many areas of mathematics and science. One common question in trigonometry is whether cos(3nπ2) can be equal to cos(n2π). This article aims to provide a comprehensive analysis of this problem, exploring the conditions under which these two expressions are equal.

Periodicity and Simplification

A key concept in trigonometry is the periodicity of the cosine function. The cosine function has a periodicity of 2π, meaning that cos(θ) cos(θ 2kπ) for any integer k. We will use this property to simplify the expressions.

Evaluating cos(3nπ/2)

First, we rewrite the angle 3nπ2 in terms of n2π:

3nπ2 n2π nπ

Since cos(θ) cos(θ 2kπ), we can simplify:

cos(3nπ2) cos(n2π nπ) cos(n2π) if n is even.

However, if n is odd:

cos(3nπ2) cos(n2π π) cos(n2π π) -cos(n2π).

Evaluating cos(nπ/2)

The value of cos(n2π) depends on the residue of n modulo 4:

If n ≡ 0 pmod{4}, then cos(n2π) 1. If n ≡ 1 pmod{4}, then cos(n2π) 0. If n ≡ 2 pmod{4}, then cos(n2π) -1. If n ≡ 3 pmod{4}, then cos(n2π) 0.

Conclusion

From the above analysis, we can draw the following conclusions:

The equality cos(3nπ2) cos(n2π) holds true only when n is even. For odd values of n, the two expressions are not equal. Specifically, cos(3nπ2) - cos(n2π).

Therefore, to determine if cos(3nπ2) cos(n2π), we need to check if n is even. If n is odd, the expressions are not equal.