Problem 35 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 5-40: Find the amplitude, the period, and the phase shift and sketch the graph of the equation. $$ y=-\sqrt{2} \sin \left(\frac{\pi}{2} x-\frac{\pi}{4}\right) $$

Step-by-Step Solution

Answer

Amplitude: \(\sqrt{2}\), Period: 4, Phase Shift: \(\frac{1}{2}\) right.

Step 1: Identify the Amplitude

The amplitude of a sine function in the form \(y = a \sin(bx + c)\) is given by \(|a|\). In this case, \(a = -\sqrt{2}\). Thus, the amplitude is \(|-\sqrt{2}| = \sqrt{2}\).

Step 2: Determine the Period

The period of a sine function is calculated using the formula \(\frac{2\pi}{|b|}\), where \(b\) is the coefficient of \(x\). Here, \(b = \frac{\pi}{2}\). Therefore, the period is \(\frac{2\pi}{\left|\frac{\pi}{2}\right|} = 4\).

Step 3: Calculate the Phase Shift

The phase shift of a sine function is determined by solving \(bx + c = 0\) for \(x\). First, rearrange \(\frac{\pi}{2}x - \frac{\pi}{4} = 0\) to get \(x = \frac{\pi/4}{\pi/2} = \frac{1}{2}\). Thus, this implies a phase shift of \(\frac{1}{2}\) units to the right.

Step 4: Sketch the Graph

To sketch the graph, start by marking the phase shift of \(\frac{1}{2}\) units to the right on the x-axis. From this point, draw one complete wave of the sine function over the period of 4 units, reaching peaks and troughs at \(\pm\sqrt{2}\), and reflecting vertically since the coefficient of the sine function is negative.

Key Concepts

Sine Function

The sine function is one of the fundamental trigonometric functions and is often symbolized as \( \sin \). It describes a smooth, periodic oscillation and is commonly used to model waves and circular motion. In its most basic form, the sine function is expressed as \( y = a \sin(bx + c) \), where each variable influences different aspects of the wave's shape:

• \(a\) adjusts the amplitude of the wave, changing how tall or deep the wave reaches.
• \(b\) affects the period, determining how frequently the wave repeats.
• \(c\) relates to the phase shift, which shifts the graph horizontally across the x-axis.

The sine function is notorious for its characteristic wave-like shape, featuring peaks and troughs that offer insights into cyclic behaviors in real-world scenarios.
To fully understand a sine function, you must analyze its amplitude, period, and phase shift, which altogether define its graphical representation.

Amplitude

Amplitude is a measure of how far the wave peaks and troughs from its central axis. It is always a positive value that shows the height of the wave. For a function of the form \( y = a \sin(bx + c) \), the amplitude is given by \( |a| \).

In the given equation \( y = -\sqrt{2} \sin \left(\frac{\pi}{2}x - \frac{\pi}{4}\right) \):

• The coefficient \( a \) is \( -\sqrt{2} \), indicating that the amplitude is \( | -\sqrt{2} | = \sqrt{2} \).

Understanding amplitude is crucial because it impacts how far the wave reaches above and below its central line. In physical contexts, this can correspond to the intensity or energy of the wave.
Moreover, a negative sign before the amplitude value, such as \( -\sqrt{2} \), introduces a reflection across the x-axis. This means the wave is inverted, swapping its peaks and troughs.

Period

The period of a sine function describes how long it takes to complete one full cycle of the wave. It is calculated using the formula \( \frac{2\pi}{|b|} \), where \( b \) is the coefficient of the \( x \) term in the sine function. This formula accounts for modifications in the wave's frequency caused by \( b \).

For the function \( y = -\sqrt{2} \sin \left(\frac{\pi}{2} x - \frac{\pi}{4} \right) \):

• The value of \( b \) is \( \frac{\pi}{2} \).
• The period is \( \frac{2\pi}{\left| \frac{\pi}{2} \right|} = 4 \).

The period indicates how often the sine wave pattern repeats itself across the x-axis.
This concept is fundamental in visualizing phenomena like sound waves or seasonal cycles, where you need to know how frequently patterns occur over time.

Phase Shift

Phase shift refers to the horizontal displacement of the sine wave along the x-axis. It shows how the wave is shifted from its usual starting position. You can find the phase shift by solving the equation \( bx + c = 0 \) for \( x \).

For the function \( y = -\sqrt{2} \sin \left(\frac{\pi}{2}x - \frac{\pi}{4}\right) \):

• Rewrite the expression \( \frac{\pi}{2}x - \frac{\pi}{4} = 0 \).
• Solve for \( x \) to get \( x = \frac{\pi/4}{\pi/2} = \frac{1}{2} \).

This results in a phase shift of \( \frac{1}{2} \) units to the right.
A phase shift essentially alters where the sine wave starts along the x-axis, which can impact how you interpret its behavior in contexts such as signal processing or oscillation analysis.