Symmetry
b. Use infinite series to show that sin x is an odd function. That is, show sin (-x) = -sin x.

Verified step by step guidance

1. Recall the infinite series expansion for \( \sin x \), which is given by: \[ \sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \]
2. To show that \( \sin x \) is an odd function, we need to evaluate \( \sin(-x) \) using the series expansion. Substitute \( -x \) into the series: \[ \sin(-x) = \sum_{n=0}^{\infty} (-1)^n \frac{(-x)^{2n+1}}{(2n+1)!} \]
3. Simplify the powers of \( -x \). Since the exponent \( 2n+1 \) is always odd, \( (-x)^{2n+1} = (-1)^{2n+1} x^{2n+1} = -x^{2n+1} \). Substitute this back into the series: \[ \sin(-x) = \sum_{n=0}^{\infty} (-1)^n \frac{-x^{2n+1}}{(2n+1)!} = - \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} \]
4. Recognize that the series inside the summation is exactly the series for \( \sin x \). Therefore, we have: \[ \sin(-x) = - \sin x \]
5. This shows that \( \sin x \) satisfies the property of an odd function, since \( \sin(-x) = -\sin x \).

Key Concepts

Odd and Even Functions

A function f(x) is odd if f(-x) = -f(x) for all x, and even if f(-x) = f(x). Understanding these definitions helps classify functions based on their symmetry about the origin or y-axis, which is essential for analyzing the sine function's behavior.

Infinite Series Representation of Sine

The sine function can be expressed as an infinite Taylor series: sin x = x - x^3/3! + x^5/5! - x^7/7! + ... . This series expansion allows us to analyze the function term-by-term, which is useful for proving properties like oddness.

Substitution in Series and Term-by-Term Analysis

By substituting -x into the sine series and examining each term, we observe how powers of -x affect the sign. Since sine's series contains only odd powers, substituting -x introduces a factor of (-1)^{odd} = -1, demonstrating sin(-x) = -sin x.