Problem 152 - Precalculus

Question

For the following exercises, find the function if \(\sin t=\frac{x}{x+1}\) $$ \tan ^{-1}\left(\frac{x}{\sqrt{2 x+1}}\right) $$

Step-by-Step Solution

Answer

The function is \( f(x) = t = \tan^{-1}\left(\frac{x}{\sqrt{2x+1}}\right) \).

Step 1: Identify the given expressions

We are given two expressions: \( \sin t = \frac{x}{x+1} \) and \( \tan^{-1} \left( \frac{x}{\sqrt{2x+1}} \right) \). We need to find a function involving these.

Step 2: Consider trigonometric identity

Recognize that given \( \sin t = \frac{x}{x+1} \), \( t \) represents an angle where sine is known. We can express \( \cos t = \sqrt{1 - \sin^2 t} \) to find cosine in terms of \( x \).

Step 3: Compute \( \cos t \)

Substitute \( \sin t \) into the identity \( \cos t = \sqrt{1 - \sin^2 t} \):\[ \cos t = \sqrt{1 - \left( \frac{x}{x+1} \right)^2 } = \sqrt{ \frac{(x+1)^2 - x^2}{(x+1)^2} } = \sqrt{\frac{2x + 1}{(x+1)^2}}.\]

Step 4: Find \( \tan t \)

Find \( \tan t \) using the identities: \( \tan t = \frac{\sin t}{\cos t} \). Substitute the known values: \[ \tan t = \frac{ \frac{x}{x+1} }{ \sqrt{\frac{2x+1}{(x+1)^2}} } = \frac{x}{\sqrt{2x+1}}. \]

Step 5: Relate to given inverse trigonometric expression

Observe that \( \tan^{-1}\left(\frac{x}{\sqrt{2x+1}}\right) \) represents angle \( t \) since \( \tan t = \frac{x}{\sqrt{2x+1}} \). Thus, the function \( f(x) \) is simply \( t \).

Step 6: Conclusion

We have found that \( f(x) = t \) where \( t \) is an angle satisfying the given trigonometric equations. So the function is \( f(x) = \tan^{-1}\left(\frac{x}{\sqrt{2x+1}}\right) \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions play an essential role in solving trigonometric equations. They help find angles when the ratio of the sides of a right triangle is known. In our exercise, we encounter the inverse tangent, denoted as \( \tan^{-1} \). Given an expression like \( \tan^{-1}\left(\frac{x}{\sqrt{2x+1}}\right) \), the inverse tangent helps us find an angle \( t \) whose tangent is \( \frac{x}{\sqrt{2x+1}} \). Understanding inverse trigonometric functions helps because:

• They allow you to find angles from known side ratios.
• They are crucial in solving equations where an angle is embedded in a function, helping transition from a ratio value back to the angle.

When computing with inverse functions, remember they often involve principal values, typically falling within a constrained range. For \( \tan^{-1} \), this range is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). This constraint ensures a single, unique value for the inverse function, making it easier to solve for specific angles in practical applications.

Trigonometric Identities

Trigonometric identities are equations true for all values of the variables involved and aid in simplifying and solving trigonometric expressions. A major identity used here is the Pythagorean identity, \( \sin^2 t + \cos^2 t = 1 \). Given \( \sin t = \frac{x}{x+1} \), we can find \( \cos t \) using this identity by implying \( \cos t = \sqrt{1 - \sin^2 t} \).Here's why trigonometric identities are beneficial:

• They simplify expressions and solve equations involving trig functions.
• They provide relationships among trigonometric functions that make understanding complex expressions easier.

Utilizing identities like \( \tan t = \frac{\sin t}{\cos t} \) makes it straightforward to compute the tangent and relate it to our inverse function problem. This use not only makes solving equations easier but also illustrates the interconnections between various trigonometric functions, providing a rich understanding of trigonometry.

Function Composition

Function composition involves applying one function to the results of another. A composed function essentially combines multiple functions into a single expression, offering deeper insights into how variables are interrelated. In trigonometry, this involves substituting one trigonometric identity into another to develop a simpler, more comprehensible equation. Navigating through function composition:

• Allows for step-by-step evaluation and simplification of complex functions.
• Enables the understanding of how manipulating outputs of one function can influence another.

In our exercise, extracting the angle \( t \) through \( \tan t = \frac{x}{\sqrt{2x+1}} \), and then using \( \tan^{-1} \) to find \( t \) as a function of \( x \), effectively demonstrates the composition of functions. Mastering this concept is crucial because it allows for handling multistage processes in mathematics, solving not just theoretical problems but also practical ones encountered in engineering and applied sciences.