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Chapter 10: Problem 8

\(f(\theta)=\cos ^{2} \theta, \quad-\pi \leq \theta \leq \pi\)

Short Answer

Expert verified
Upon analyzing the function, the graph of \(f(\theta)=\cos ^{2} \theta\) over the range \(-\pi \leq \theta \leq \pi\) is found to be symmetric about the y-axis with a period of \(\pi\). The graph can be sketched using these properties along with the calculated values at some fundamental points.

Step by step solution

01

Identify Key Values

First it is valuable to identify a few key points. Since the function is \(f(\theta)=\cos ^{2} \theta\), take the square of cosine for some fundamental points: \[\theta=-\pi, -\pi/2, 0, \pi/2, \pi\] Then calculate and record these values.
02

Identify Symmetry

Next, use the cosine function's symmetry properties to help sketch this graph. Due to the nature of cosine, this function should be symmetric about the y-axis, meaning it should look the same if flipped left-to-right.
03

Identify Period

Another factor is the period of the function. The period of the base cosine function is \(2\pi\), but since we are squaring it, the resulting period for the function \(f(\theta)=\cos ^{2} \theta\) is \(\pi\). This means that the function will repeat itself after every interval of length \(\pi\). This will also be useful in sketching.
04

Sketch the Function

With the identified points (Step 1), the symmetry (Step 2), and the function's period (Step 3), sketch the function over the interval \(-\pi \leq \theta \leq \pi\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Cosine Squared Function
When we talk about the cosine squared function, we refer to the square of the cosine trigonometric function. Mathematically, it is expressed as \( f(\theta) = \cos^2 \theta \). This function takes the cosine of an angle \( \theta \) and then squares the result.

To visualize this function, we can start by plotting the basic cosine function, \( \cos \theta \), and then simply squaring each of its output values. The result is a function that no longer dips below the horizontal axis because squaring a number鈥攚hether it's positive or negative鈥攁lways gives a non-negative result. Linked to the exercise provided, when calculating key values--like where \( \theta = -\pi, -\pi/2, 0, \pi/2, \pi \)--we can see that the squaring effect alters the typical wave pattern of the cosine function, creating a graph that touches the horizontal axis at the peaks of the original cosine function, but never below.
Trigonometric Function Symmetry
Symmetry in trigonometric functions is a fundamental aspect that can simplify the understanding and sketching of these functions. For the cosine squared function, symmetry is an especially useful property. Cosine functions are even functions, which means they are symmetric about the y-axis.

When you square the cosine function to get \( f(\theta) = \cos^2 \theta \), it retains this symmetry. This can be observed by noting that \( \cos(\theta) \) is equal to \( \cos(-\theta) \) for any angle \( \theta \), which means that their squares are also equal. Consequently, if you plot the positive side of the function, you can reflect the curve about the y-axis to plot the negative side. This cuts the work in half when sketching the graph by hand, as highlighted in Step 2 of the exercise.
Trigonometric Function Period
The concept of the period of a trigonometric function defines the length of the smallest interval over which the function's values repeat. For the base cosine function, this period is \( 2\pi \), which means that every \( 2\pi \) units along the horizontal axis, the function's wave pattern starts again.

However, when we deal with the cosine squared function, the period is actually halved to \( \pi \) as shown in Step 3 of the solution because the squaring effect causes the function to complete a full wave of its pattern over a shorter interval. This is a significant concept to grasp as it impacts how you sketch the function and can help with identifying patterns and repetitions in trigonometric graphs.

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Most popular questions from this chapter

$$ \frac{\partial^{2} u}{\partial t^{2}} $$ \(=\frac{\partial^{2} u}{\partial x^{2}}+x \sin t\), \(0\)<\(x\)<\(\pi\), \(t>0\) $$ u(0, t) $$ \(=u(\pi, t)=0\) \(t>0\), $$ u(x, 0)=0 $$, \(0\)<\(x\)<\(\pi\), $$ \frac{\partial u}{\partial t}(x, 0) $$ \(=0\), \(0\)<\(x\)<\(\pi\)

$$\frac{\partial u}{\partial t}=2 \frac{\partial^{2} u}{\partial x^{2}}, 0<\mathcal{x}<\pi,t>0,$$ $$u(0, t)=5, \quad u(\pi, t)=10, \quad t>0,$$ $$u(x, 0)=\sin 3 x-\sin 5 x,0<\mathcal{x}<\pi$$

A vibrating string is governed by the initial-boundary value problem \((1)-(4)\) . The initial conditions are given by \(f(x)=0\) and \(g(x)=\left\\{\begin{array}{l}{v_{0} x / a} \\ {v_{0}(L-x) /(L-a)}\end{array}\right.\) \(0<\)x \(\leq a\) \(a<\)x<\(L\) where \(v_{0}\) is a constant. Find a formal solution.

Unbounded Domain. Using separation of variables, find a solution of Laplace's equation in the infinite rectangle \(0< x <\pi, 0< y <\infty\) that is zero on the sides \(x=0\) and \(x=\pi,\) approaches zero as \(y \rightarrow \infty,\) and equals \(f(x)\) for \(y=0.\)

$$f(x)=x$$ \(-\pi<\) \(x<\pi\)
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