Problem 38 - Precalculus

Question

Prove or disprove the identity. $$\frac{\sec (-x)}{\tan x+\cot x}=-\sin (-x)$$

Step-by-Step Solution

Answer

The identity is correct.

Step 1: Simplify the left side

We start by examining the left side of the equation \(\frac{\sec(-x)}{\tan x + \cot x}\). The trigonometric identity \(\sec(-x) = \sec x\) holds because secant is an even function. Thus, we have \(\frac{\sec x}{\tan x + \cot x}\).

Step 2: Rewrite tangent and cotangent

Now, express \(\tan x\) and \(\cot x\) in terms of sine and cosine: \(\tan x = \frac{\sin x}{\cos x}\) and \(\cot x = \frac{\cos x}{\sin x}\). Thus, \(\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\).

Step 3: Simplify the denominator

Combine the fractions in the denominator: \(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x}\). Recall that \(\sin^2 x + \cos^2 x = 1\). Therefore, the denominator simplifies to \(\frac{1}{\sin x \cos x}\).

Step 4: Simplify the fraction

Now, simplify the entire expression: \[\frac{\sec x}{\frac{1}{\sin x \cos x}} = \sec x \cdot \sin x \cos x\]. Replace \(\sec x\) with \(\frac{1}{\cos x}\) to get \[\frac{1}{\cos x} \cdot \sin x \cos x = \sin x\].

Step 5: Simplify the right side

Examine the right side \(-\sin(-x)\). Knowing that sine is an odd function, we have \(-\sin(-x) = \sin x\).

Step 6: Compare both sides

After simplification, the left side \(\sin x\) matches the right side \(\sin x\). Therefore, the original identity \(\frac{\sec(-x)}{\tan x + \cot x} = -\sin(-x)\) is correct.

Key Concepts

Even and Odd Functions

Understanding even and odd functions can be really helpful when dealing with trigonometric equations. An even function is one that satisfies the condition:

• \( f(-x) = f(x) \)

This means that their graph is symmetric with respect to the y-axis. Common examples of even functions in trigonometry include cosine and secant.
On the other hand, odd functions satisfy the condition:

• \( f(-x) = -f(x) \)

This means their graph is symmetric with respect to the origin. Examples include sine, tangent, and cotangent.
In our exercise, we use the property that secant is an even function: \( \sec(-x) = \sec x \) and sine is an odd function: \( \sin(-x) = -\sin x \). These properties allow us to simplify the expression and find equivalences in trigonometric identities.

Simplifying Fractions

Simplifying fractions is a crucial skill in mathematics. It allows us to break down and understand complex expressions more easily.
When simplifying fractions, we should aim to write the expression in its simplest form. Generally, this means combining like terms and canceling out common factors in the numerator and denominator.
In trigonometric contexts, this often involves using identities like those for sine, cosine, tangent, and cotangent. For instance:

• We can express \( \tan x \) as \( \frac{\sin x}{\cos x} \)
• We can express \( \cot x \) as \( \frac{\cos x}{\sin x} \)

By rewriting these terms, we can combine and simplify fractions effectively. In our problem, we combined the fractions \( \frac{\sin x}{\cos x} \) and \( \frac{\cos x}{\sin x} \) to get \( \frac{1}{\sin x \cos x} \). This simplification hinges on the identity \( \sin^2 x + \cos^2 x = 1 \).

Trigonometric Functions

Trigonometric functions are fundamental in understanding the relationships of angles within triangles and a circle. Here's a brief overview of the core functions:

• Sine (\(\sin\)): \( \sin x = \frac{\text{opposite}}{\text{hypotenuse}} \)
• Cosine (\(\cos\)): \( \cos x = \frac{\text{adjacent}}{\text{hypotenuse}} \)
• Tangent (\(\tan\)): \( \tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin x}{\cos x} \)
• Cotangent (\(\cot\)): \( \cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x} \)
• Secant (\(\sec\)): \( \sec x = \frac{1}{\cos x} \)
• Cosecant (\(\csc\)): \( \csc x = \frac{1}{\sin x} \)

These functions have specific properties like being even or odd, which help simplify expressions. In solving our exercise, recognizing that \(\sec(-x)\) and \(\sin(-x)\) behave predictably due to their properties as even and odd functions, respectively, greatly aids in proving or disproving identities.