/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 Exercises \(64-66\) will help yo... [FREE SOLUTION] | 91Ó°ÊÓ

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Exercises \(64-66\) will help you prepare for the material covered in the next section. In each exercise, complete the table of coordinates. Do not use a calculator. $$ y=\frac{1}{2} \cos (4 x+\pi) $$ $$ \begin{aligned} &\begin{array}{lllll} {x} & {-\frac{\pi}{4}} & {-\frac{\pi}{8}} & {0} & {\frac{\pi}{8}} & {\frac{\pi}{4}} \end{array}\\\ &y \end{aligned} $$ After completing this table of coordinates, plot the nine ordered pairs as points in a rectangular coordinate system. Then connect the points with a smooth curve.

Short Answer

Expert verified
The coordinates after evaluating the function at the given x-values are (-Ï€/4, 1/2), (-Ï€/8, 0), (0, -1/2), (Ï€/8, 0), and (Ï€/4, 1/2)

Step by step solution

01

Evaluating function at x=-Ï€/4

Substitute \(x = -π/4\) into the equation. So it will be \(y = 1/2 cos(4*(-π/4) + π) = 1/2 cos(-π + π) = 1/2 cos(0) = 1/2\). So the coordinate at \(x = -π/4\) is \((-π/4, 1/2)\)
02

Evaluating function at x=-Ï€/8

Substitute \(x = -π/8\) into the equation. So we get \(y = 1/2 cos(4*(-π/8) + π) = 1/2 cos(-π/2 + π) = 1/2 cos(π/2) = 0\). So the coordinate at \(x = -π/8\) is \((-π/8, 0)\)
03

Evaluating function at x=0

Substitute \(x = 0\) into the equation. That gives \(y = 1/2 cos(4*0 + π) = 1/2 cos(π) = -1/2\). So the coordinate at \(x = 0\) is \((0, -1/2)\)
04

Evaluating function at x=Ï€/8

Substitute \(x = π/8\) into the equation. So it will be \(y = 1/2 cos(4*(π/8) + π) = 1/2 cos(π/2 + π) = 1/2 cos(3π/2) = 0\). So the coordinate at \(x = π/8\) is \((π/8, 0)\)
05

Evaluating function at x=Ï€/4

Substitute \(x = π/4\) into the equation. So it will be \(y = 1/2 cos(4*(π/4) + π) = 1/2 cos(π + π) = 1/2 cos(2π) = 1/2\). So the coordinate at \(x = π/4\) is \((π/4, 1/2)\)
06

Plotting on a rectangular coordinate system

Mark all the points on a coordinate grid. Specifically, plot (-Ï€/4, 1/2), (-Ï€/8, 0), (0, -1/2), (Ï€/8, 0), and (Ï€/4, 1/2). Then connect all of the plotted points smoothly.

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Key Concepts

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

Coordinate System
The coordinate system is a framework used to determine the position of points in a plane. It consists of two perpendicular lines or axes: the horizontal axis (x-axis) and the vertical axis (y-axis). These axes intersect at the origin, denoted as the point \(0, 0\). Each point on this plane is defined by an ordered pair of numbers \(x, y\), which represent its location relative to these two axes.

In a rectangular coordinate system, which is often referred to in mathematics as a Cartesian coordinate system, coordinates are precise and easy to understand. This system allows us to graphically represent various mathematical functions and their relationships. In this context, our primary interest is in plotting trigonometric functions such as the cosine function. To do this, we need to know both the x-value and the y-value of each point. When plotting functions, the x-values typically represent inputs into the function, and the resulting y-values are the outputs.
Cosine Function
The cosine function is one of the fundamental trigonometric functions used to describe the relationship between the angles and sides of a right triangle. It is often denoted as \cos(\theta)\, where \theta\ is the angle. The cosine function is periodic with a period of \(2\pi\), meaning it repeats its values every \(2\pi\) radians.

In our specific example, we explore a modified version of the cosine function: \(y = \frac{1}{2} \cos(4x + \pi)\). Here, a few modifications are clear:
  • Amplitude: The \(\frac{1}{2}\) outside the cosine function scales the amplitude of the wave, halving the typical peak and trough values of -1 to 1, making them -1/2 to 1/2.
  • Frequency: The coefficient 4 in front of \(x\) increases the frequency, so the function completes its cycle four times as fast as the standard cosine function.
  • Phase Shift: The \(+ \pi\) inside the function results in a horizontal shift in the function. Specifically, this translates the graph by matching the calculated phase shift component.
Understanding these changes is crucial for accurately plotting the modified cosine function.
Graphing Trigonometric Functions
Graphing trigonometric functions is an essential skill for visualizing how these mathematical curves behave over a given interval. When graphing a specific function like \(y = \frac{1}{2} \cos(4x + \pi)\), you'll need to examine key properties such as amplitude, period, frequency, and any shifts.

Start by calculating a few crucial points along the curve, as per the exercise instructions. Each x value results in a corresponding y value when plugged into the equation. These calculated pairs \((x, y)\) are then plotted on the rectangular coordinate system:
  • For example, when \(x = 0\), the calculation gives us \(y = -1/2\).
  • When \(x = \frac{\pi}{4}\), it results in \(y = 1/2\), and so forth for other specified x values.
Connect these plotted points with a smooth, continuous curve. By doing so, you visually represent the cyclical nature of the trigonometric function you are examining.
Evaluating Trigonometric Expressions
Evaluating trigonometric expressions involves substituting values into the function and calculating the corresponding output. To find y-values for a function like \(y = \frac{1}{2} \cos(4x + \pi)\), each x-value must be substituted into the expression:

  • For \(x = -\frac{\pi}{4}\), substitute into the equation to find \(y = \frac{1}{2} \cos\big(-\pi + \pi\big) = \frac{1}{2} \cos 0 = \frac{1}{2}\).
  • Continue this process for each x-value given: substitute, simplify, and solve for y.
Understanding how to manually evaluate these expressions without a calculator can strengthen your comprehension of trigonometric functions and help in visualizing their graphs. Trigonometric identities and the unit circle are often leveraged to find these values accurately.

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Most popular questions from this chapter

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