Problem 116 - Algebra and Trigonometry

Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. If \(f(\theta)=\sqrt{25-\theta^{2}}\) and \(g(\theta)=5 \sin \theta,-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\) show that \((f \circ g)(\theta)=5 \cos \theta\)

Step-by-Step Solution

Answer

(f \circ g)(\theta) = 5 \cos \theta

Step 1: Understand Given Functions

Given the functions: 1. \(f(\theta)=\sqrt{25-\theta^{2}}\) 2. \(g(\theta)=5 \sin \theta\) with \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \)

Step 2: Composition of Functions

Find the composition \(f \circ g\)(\theta), which is defined as \(f(g(\theta))\). Here, replace \(\theta\) in \(f(\theta)\) with \(g(\theta)\).

Step 3: Substitute \(g(\theta)\) into \(f(\theta)\)

Since \(g(\theta) = 5 \sin \theta\), substitute this into \(f\): \[ f(g(\theta)) = \sqrt{25 - (5 \sin \theta)^2} \]

Step 4: Simplify the Expression

Simplify the expression inside the square root: \[ 25 - (5 \sin \theta)^2 = 25 - 25 \sin^2 \theta \] Now we have: \[ \sqrt{25 - 25 \sin^2 \theta} = \sqrt{25(1 - \sin^2 \theta)} \]

Step 5: Use Trigonometric Identity

Recall the trigonometric identity: \( \sin^2 \theta + \cos^2 \theta = 1 \), which gives: \(1 - \sin^2 \theta = \cos^2 \theta\). Using this, simplify to: \[ \sqrt{25 \cos^2 \theta} \]

Step 6: Simplify Further

Since \(\sqrt{a^2} = |a|\), and knowing \( \cos \theta \) is non-negative in the given interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\), simplify to: \[ \sqrt{25 \cos^2 \theta} = 5 \cos \theta \]

Step 7: Conclude the Result

Thus, we have shown that: \[ (f \circ g)(\theta) = 5 \cos \theta \]

Key Concepts

function composition

Function composition is a crucial concept in mathematics that combines two functions to create a third function. In general, if you have two functions, say \( f \) and \( g \), the composition of these functions is written as \( f(g(x)) \). This means you take the output from function \( g \) and use it as the input for function \( f \). Think of function composition as a process where an item moves through a sequence of functions.

For example, in the given exercise, we have \( f(\theta) = \sqrt{25 - \theta^2} \) and \( g(\theta) = 5 \sin \theta \). To find \((f \circ g)(\theta)\), we substitute \(g(\theta)\) into \(f(\theta)\). So, we get \( f(g(\theta)) = \sqrt{25 - (5 \sin \theta)^2} \).

This step-by-step substitution process helps break down complex functions into more manageable parts, improving our understanding and making calculations easier.

trigonometric identities

Trigonometric identities are equations that relate the angles of a triangle to its side lengths. These identities are fundamental in simplifying expressions and solving trigonometric equations.

In our exercise, the key trigonometric identity used is \( \sin^2 \theta + \cos^2 \theta = 1 \). This identity facilitates the simplification of the expression inside the square root. For instance, when we have \(25(1 - \sin^2 \theta)\), we use the identity \(1 - \sin^2 \theta = \cos^2 \theta\).

Understanding these identities is essential as they appear frequently in problems involving angles and can significantly simplify complex expressions. They help in converting one trigonometric function to another, making problem-solving more efficient.

interval analysis

Interval analysis is the study of the range of values that a function can take over a specific interval. It helps us understand the behavior of functions within given boundaries.

In the exercise, the interval for \( \theta \) is given as \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \). This interval is crucial because it tells us that \( \theta \) ranges from \(-90^\text{o} \) to \( 90^\text{o} \).

Since \( \sin(\theta) \) and \( \cos(\theta) \) are bounded within this interval, it ensures that the values of \( 5 \sin \theta \) and \( 25 - 25 \sin^2 \theta \) are well-defined. Moreover, knowing that \( \cos(\theta) \) is non-negative in this interval ensures that our final result, \( 5 \cos \theta \), is valid and meets the criteria set by the problem.

square root simplification

Square root simplification involves reducing expressions under the square root to simpler forms. This simplification makes it easier to understand and solve mathematical problems.

In our exercise, we encounter the nested expression \( \sqrt{25(1 - \sin^2 \theta)} \). The process involves:

• First, substituting \(1 - \sin^2 \theta\) with \(\cos^2 \theta\) using the trigonometric identity.
• Then, simplifying \( 25 \cos^2 \theta \) to \(5^2 \cos^2 \theta\).

Finally, since the square root of \( a^2 \) is \( |a| \), and within the given interval, \( \cos(\theta) \) is non-negative, we simplify to \( 5 \cos \theta \).

This step ensures that we have fully simplified the function, making it straightforward and easier to handle in subsequent calculations. Simplifying expressions under square roots is a vital skill, especially when dealing with more complex mathematical functions.