Problem 1

Question

Since $f(-x)=\sin (-3 x)=-\sin 3 x=-f(x), f(x)$ is an odd function.

Step-by-Step Solution

Verified

Answer

The function $ f(x) = \sin(-3x) $ is an odd function because it satisfies $ f(-x) = -f(x) $.

1Step 1: Identify the Function

The exercise gives that $ f(x) = \sin(-3x) $. Our goal is to determine whether this function is odd or even.

2Step 2: Understand Odd and Even Functions

An odd function satisfies $ f(-x) = -f(x) $. An even function satisfies $ f(-x) = f(x) $. We will verify if $ f(x) = \sin(-3x) $ fits one of these properties.

3Step 3: Substitute -x into the Function

Substitute $-x$ into the function: $ f(-x) = \sin(-3(-x)) = \sin(3x) $.

4Step 4: Compare $ f(-x) $ and $ f(x) $

From Step 3, we have $ f(-x) = \sin(3x) $. The given statement says $ f(x) = \sin(-3x) $. We need to verify if the equality $ \sin(3x) = -\sin(-3x) $ is true.

5Step 5: Use Trigonometric Identity

Recall the trigonometric identity $ \sin(-\theta) = -\sin(\theta) $. So, $ \sin(3x) = -\sin(-3x) $ becomes $ \sin(3x) = -(-\sin(3x)) $, simplifying to $ \sin(3x) = \sin(3x) $.

6Step 6: Conclude the Function's Property

Since $ f(-x) = -f(x) $ holds true, the function $ \sin(-3x) $ satisfies the condition for being an odd function.

Key Concepts

Trigonometric IdentityEven FunctionsFunction Properties

Trigonometric Identity

Trigonometric identities are equations that involve trigonometric functions and are valid for every value of the occurring variables where both sides are defined. These identities are incredibly useful in simplifying complex trigonometric expressions and proving other mathematical statements. A popular trigonometric identity, crucial for this problem, is the odd identity of the sine function:

$ \sin(-\theta) = -\sin(\theta) $

This identity indicates that the sine function changes its sign when the angle is negated, which hints at the symmetric property of the sine function with respect to the origin. In our exercise, using $ \sin(-3x) = -\sin(3x) $ helped us determine the nature of the function $ f(x) = \sin(-3x) $ by showing that the function is not even, but odd. Understanding and applying these identities can make solving functions involving trigonometry more intuitive and manageable.

Even Functions

Even functions are characterized by their symmetry about the y-axis. If you fold the graph of an even function along the y-axis, both halves would match perfectly, just like mirror images. The mathematical description of even functions is that:

$ f(-x) = f(x) $

This means that if you input the negative of any number into the function, you will get the same value as if you had input the original number. Common examples of even functions include the cosine function, where $ \cos(-x) = \cos(x) $, and quadratic functions like $ f(x) = x^2 $. However, it is important to note that not all functions are either even or odd; they can be neither. By examining whether the given function $ f(x) = \sin(-3x) $ satisfies the condition for even functions, we can better categorize the function and understand its behavior.

Function Properties

To fully analyze the properties of a function, we must consider various aspects, such as symmetry, periodicity, and continuity. In trigonometric functions, symmetry is one of the key properties used to classify them as either odd or even.With odd functions, like our example, symmetry is about the origin. This means that rotating the function graph by 180 degrees around the origin results in an identical graph. Mathematically, for a function to be classified as an odd function:

$ f(-x) = -f(x) $

The trigonometric function $ \sin(x) $ is a prime example, as it follows this rule, confirming its odd nature. Understanding these function properties is fundamental as they allow us to predict and describe the behavior of functions across different inputs and scenarios. Recognizing the odd nature of $ f(x) = \sin(-3x) $ based on its properties helps in graphing and transforming them appropriately for various applications.

Problem 1

Other exercises in this chapter

Problem 1

Identifying $p=2$ we have $$\begin{aligned} c_{n} &=\frac{1}{4} \int_{-2}^{2} f(x) e^{-i n \pi x / 2} d x=\frac{1}{4}\left[\int_{-2}^{0}(-1) e^{-i n \pi x / 2

View solution

Problem 1

For $\lambda \leq 0$ the only solution of the boundary-value problem is $y=0 .$ For $\lambda=\alpha^{2}>0$ we have $$y=c_{1} \cos \alpha x+c_{2} \sin \alp

View solution

Problem 1

$$\int_{-2}^{2} x x^{2} d x=\left.\frac{1}{4} x^{4}\right|_{-2} ^{2}=0$$

View solution

Problem 2

Since $f(-x)=-x \cos (-x)=-x \cos x=-f(x), f(x)$ is an odd function.

View solution