Problem 76 - Precalculus Mathematics for Calculus

Question

Determine whether the function is even, odd, or neither. $$f(x)=\sin x+\cos x$$

Step-by-Step Solution

Answer

The function \( f(x) = \sin x + \cos x \) is neither even nor odd.

Step 1: Understand the Definition of Even and Odd Functions

An even function satisfies the condition that \( f(-x) = f(x) \) for all \( x \). An odd function satisfies the condition \( f(-x) = -f(x) \) for all \( x \). If a function satisfies neither condition, it is neither even nor odd.

Step 2: Substitute \(-x\) into the Function

To test if the function is even or odd, substitute \(-x\) into the function: \( f(-x) = \sin(-x) + \cos(-x) \). Using trigonometric identities: \( \sin(-x) = -\sin(x) \) and \( \cos(-x) = \cos(x) \).

Step 3: Simplify the Expression

Now, simplify \( f(-x) \) using the identities: \( f(-x) = -\sin(x) + \cos(x) \).

Step 4: Compare \( f(-x) \) to \( f(x) \)

Compare \( f(-x) = -\sin(x) + \cos(x) \) with \( f(x) = \sin(x) + \cos(x) \). They are not equal, so the function is not even.

Step 5: Check if the Function is Odd

To check if the function is odd, see if \( f(-x) = -f(x) \). Substitute \( -f(x) \) to get \( -\sin(x) - \cos(x) \). As \( f(-x) \) does not equal \(-f(x)\), the function is not odd.

Step 6: Determine that the Function is Neither

Since \( f(x) \) is not even (\( f(-x) eq f(x) \)) or odd (\( f(-x) eq -f(x) \)), the function \( f(x) = \sin(x) + \cos(x) \) is neither even nor odd.

Key Concepts

Even Functions

Even functions have a unique quality. They are symmetrical relative to the vertical y-axis. This means that if you draw a graph of an even function, you can fold it along the y-axis and both halves will match perfectly. The core condition for a function to be classified as even is that its equation satisfies: \( f(-x) = f(x) \) for every value of \( x \).

Here's how it works:

• Start with the original equation \( f(x) \).
• Replace \( x \) with \( -x \) and simplify.
• If the result is the same as the original function, then \( f(x) \) is even.

Even functions can include basic polynomial functions like \( x^2 \), because \( (-x)^2 = x^2 \). For trigonometric functions, the cosine function \( \cos(x) \) is an example of an even function since \( \cos(-x) = \cos(x) \).

Odd Functions

Odd functions also possess a special symmetry, but unlike even functions, their symmetry is around the origin. This means that if you rotate the graph 180 degrees around the origin, it will look the same. For a function to be identified as odd, the equation must satisfy the condition: \( f(-x) = -f(x) \) for every \( x \).

Here's the procedure to check:

• Take the original equation \( f(x) \).
• Substitute \( -x \) in place of \( x \) and simplify.
• If the new expression is the negative version of the original, then the function is odd.

An example is the sine function \( \sin(x) \), because \( \sin(-x) = -\sin(x) \). Identifying whether a function is odd helps in understanding its behavior and symmetry, especially dealing with trigonometric problems or Fourier series.

Trigonometric Identities

Trigonometric identities are fundamental tools in simplifying and solving trigonometric equations. They are mathematical equations that are true for every value of the occurring variables. Key identities help transform and simplify expressions involving trigonometric functions.

Some core trigonometric identities include:

• Reciprocal identities, such as \( \sin \theta = \frac{1}{\csc \theta} \).
• Pythagorean identities, like \( \sin^2\theta + \cos^2\theta = 1 \).
• Even-Odd identities, which include \( \sin(-\theta) = -\sin \theta \) and \( \cos(-\theta) = \cos \theta \).

These identities allow us to convert complex trigonometric expressions into simpler forms. For instance, the even-odd identities are used in assessing whether a trigonometric function is even, odd, or neither. Understanding how and when to use these identities makes solving many types of math problems more accessible, whether you are dealing with academic exercises or real-world applications.