Problem 72 - Precalculus Mathematics for Calculus

Question

Verify the identity. $$\frac{\tan v \sin v}{\tan v+\sin v}=\frac{\tan v-\sin v}{\tan v \sin v}$$

Step-by-Step Solution

Answer

The given identity is verified.

Step 1: Understand the Identity

The given identity is \( \frac{\tan v \sin v}{\tan v+\sin v} = \frac{\tan v-\sin v}{\tan v \sin v} \). Our goal is to show that both sides are equivalent.

Step 2: Rewrite Tangent in Terms of Sine and Cosine

Recall that \( \tan v = \frac{\sin v}{\cos v} \). Substitute this into the identity: Left-hand side: \( \frac{\left(\frac{\sin v}{\cos v}\right) \sin v}{\frac{\sin v}{\cos v} + \sin v} \) Right-hand side: \( \frac{\frac{\sin v}{\cos v} - \sin v}{\left(\frac{\sin v}{\cos v}\right) \sin v} \).

Step 3: Simplify Both Sides

Left-hand side: \[ \frac{\frac{\sin^2 v}{\cos v}}{\frac{\sin v + \sin v \cos v}{\cos v}} = \frac{\sin^2 v}{\sin v + \sin v \cos v} = \frac{\sin v}{1 + \cos v} \]Right-hand side: \[ \frac{\frac{\sin v - \sin v \cos v}{\cos v}}{\frac{\sin^2 v}{\cos v}} = \frac{\sin v (1 - \cos v)}{\sin^2 v} = \frac{1 - \cos v}{\sin v} \].

Step 4: Further Manipulate to Common Form

The expressions \( \frac{\sin v}{1 + \cos v} \) and \( \frac{1 - \cos v}{\sin v} \) look different, but by rationalization or manipulating we can relate them. NoticeMultiply \( \frac{\sin v}{1+\cos v} \) by \( \frac{1-\cos v}{1-\cos v} \), which is \( 1 \): \[ \frac{\sin v (1-\cos v)}{(1 + \cos v)(1 - \cos v)} = \frac{\sin v (1 - \cos v)}{1 - \cos^2 v} \]Since \( 1 - \cos^2 v = \sin^2 v \), we get:\[ \frac{\sin v (1 - \cos v)}{\sin^2 v} = \frac{1 - \cos v}{\sin v} \].

Step 5: Confirm Identity

Both sides of the identity are now in the form \( \frac{1 - \cos v}{\sin v} \), confirming that \( \frac{\tan v \sin v}{\tan v+\sin v} = \frac{\tan v-\sin v}{\tan v \sin v} \) is true.

Key Concepts

Simplifying Trigonometric Expressions

Simplifying trigonometric expressions involves rewriting them into a more manageable form, often using trigonometric identities. This is an essential skill in solving trigonometric equations or verifying identities. In the given problem, we need to simplify both sides of the given identity to prove they are equal.

Let's first look at what simplification often involves:

• Substitution: Replacing trigonometric functions with equivalent expressions using identities can unveil hidden relationships.
• Combining fractions: Manipulating fractions or rational expressions to have a common denominator can simplify the expression.
• Factoring: Simplifying by factoring common elements in the numerator and the denominator.
• Cancellation: Canceling out terms or expressions on both sides to arrive at a simplified form.

In this particular exercise, simplifying helps us transform seemingly different expressions to reveal underlying equivalence. For instance, utilizing the identity \( 1 - \cos^2 v = \sin^2 v \) allowed us to simplify the expression further by rewriting products into a straightforward form. Remember, the goal of simplification is not only to solve but to understand the relationship between elements of the expression.

Tangent Function

The tangent function, often represented as \( \tan v \), is a fundamental trigonometric function defined as the ratio of the sine and cosine functions. Specifically, \( \tan v = \frac{\sin v}{\cos v} \). This identity is crucial when simplifying expressions because it allows us to rewrite tangent in terms of sine and cosine, which can be manipulated easily.

In our exercise, rewriting the tangent function helps break down the complex identity into simpler components. By substituting \( \tan v = \frac{\sin v}{\cos v} \) into our expression, we can factor out common terms and reveal equivalences that aren't immediately obvious.

Important properties of the tangent function include:

• Periodicity: The tangent function has a period of \( \pi \), meaning it repeats every \( \pi \) units.
• Undefined Points: \( \tan v \) is undefined when \( \cos v = 0 \), such as at odd multiples of \( \frac{\pi}{2} \).

The ability to convert among trigonometric functions using their foundational identities opens up new pathways in problem-solving and verification of identities.

Sine Function

The sine function, denoted by \( \sin v \), is one of the core trigonometric functions. The sine of an angle in a right triangle is defined as the length of the opposite side divided by the length of the hypotenuse. It is a periodic function with a period of \( 2\pi \).

In trigonometric identities, the sine function often interacts with the cosine function, as the two are closely related through identities such as \( \sin^2 v + \cos^2 v = 1 \). In this exercise, the sine function is central to simplifying the given trigonometric expression as it appears in both the numerator and denominator.

Key points to remember about the sine function:

• Range: The values of \( \sin v \) lie between -1 and 1.
• Zeros: \( \sin v = 0 \) for multiples of \( \pi \), including 0, \( \pi \), and \( 2\pi \).
• Co-function Identity: \( \sin(\pi/2 - v) = \cos v \).

Understanding the roles of sine in identities and expressions, as well as its interactions with other functions and the unit circle, is vital for both simplifying and solving trigonometric problems.