Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\quad g(y)=\cos 2 y ; \quad[0, \pi]\)

Step-by-Step Solution

Answer

The range of \( g(y) = \cos 2y \) over \([0, \pi]\) is \([-1, 1]\).

Step 1: Understand the Function Domain

The given function is \( g(y) = \cos 2y \) and its domain is \( [0, \pi] \). This means we have to consider the values of \( y \) starting from \( 0 \) up to \( \pi \).

Step 2: Determine the Range of \( 2y \)

Since the original domain of \( y \) is \( [0, \pi] \), the range of \( 2y \) becomes \( [0, 2\pi] \) after multiplying each element in the domain by 2.

Step 3: Evaluate \( \cos 2y \) over \( [0, 2\pi] \)

The cosine function \( \cos x \) over the interval \( [0, 2\pi] \) completes one full cycle. The function \( \cos x \) achieves its maximum value of \( 1 \) at \( x = 0, 2\pi \) and its minimum value of \(-1\) at \( x = \pi \).

Step 4: Determine \( g(y) \) values over \( [0, \pi] \)

Since the range of \( 2y \) is a complete cycle, the function \( g(y) = \cos 2y \) will take on all the values from \(-1\) to \( 1\).

Step 5: Conclude the Range of \( g(y) \)

The range of \( g(y) = \cos 2y \) for \( y \in [0, \pi] \) is \([-1, 1]\) because as \( y \) changes from \( 0 \) to \( \pi \), \( 2y \) goes from \( 0 \) to \( 2\pi \), where \( \cos x \) covers its full cycle.

Key Concepts

Cosine Function

The cosine function, often denoted as \( \cos x \), is a fundamental trigonometric function. It describes the relationship between the angle in a right triangle and its adjacent side over the hypotenuse. This function is periodic, which means it repeats its values in regular intervals. The cosine function repeats every \(2\pi\), known as its period.

- At \(x = 0\) and \(x = 2\pi\), \(\cos x\) equals 1.- At \(x = \pi/2\) and \(x = 3\pi/2\), its value is 0.- At \(x = \pi\), \(\cos x\) equals -1.

The cosine wave starts at its maximum value of 1, decreases to -1, and returns to 1, completing a full cycle. Understanding how the cosine function behaves across different intervals, especially \([0, 2\pi]\), helps us to comprehend its application in various mathematical problems like the given function \(g(y) = \cos 2y\).

Function Domain and Range

The domain of a function refers to all possible inputs for the function, while the range is all possible outputs. For our function \(g(y) = \cos 2y\), the domain is given as \([0, \pi]\). This means that \(y\) can take any value between 0 and \(\pi\).

Since \(g(y)\) is defined as \(\cos 2y\), to find the effect on the domain, we multiply each element of \([0, \pi]\) by 2, resulting in a new range for \(2y\), which is \([0, 2\pi]\). This expanded domain allows us to consider how \(\cos 2y\) behaves over its periodic cycle.

The range of a cosine function is always \([-1, 1]\), and since \(2y\) spans a full cycle over \([0, 2\pi]\), the function value \(g(y) = \cos 2y\) over the interval \([0, \pi]\) will also range from -1 to 1.

Function Evaluation

Function evaluation involves determining the output of a function for specific inputs. In our case, evaluating the function \(g(y) = \cos 2y\) over its domain \([0, \pi]\) involves understanding how the cosine values progress as \(y\) changes.

Let's examine the cycle of \(\cos 2y\):- At \(y = 0\), \(2y = 0\), and therefore \(\cos 2y = \cos 0 = 1\).- As \(y\) approaches \(\pi/2\), \(2y = \pi\), and \(\cos 2y = \cos \pi = -1\).- Finally, at \(y = \pi\), \(2y = 2\pi\), bringing \(\cos 2y\) back to \(\cos 2\pi = 1\).

We can see that as \(y\) moves from 0 to \(\pi\), \(\cos 2y\) recreates a full cycle of the cosine wave, providing insights into the evaluation of functions along specific domains.

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and hold true for all values of the variable where they are defined. These are essential tools in simplifying expressions and solving trigonometric equations.

For instance, one fundamental identity is the Pythagorean identity, \(\cos^2 x + \sin^2 x = 1\). This identity is incredibly useful in converting between sine and cosine functions.

Additionally, the double angle identity \(\cos 2x = \cos^2 x - \sin^2 x\) or \(\cos 2x = 2\cos^2 x - 1\) can be particularly helpful when dealing with expressions like \(\cos 2y\), which occurs in our function \(g(y) = \cos 2y\).

These identities not only aid in simplifying expressions but also help in evaluating trigonometric functions over specific intervals, understanding their behavior, and solving complex trigonometric problems efficiently.