Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 4 - Graphs of the Circular Functions - Section 4.1 Graphs of the Sine and Cosine Functions - 4.1 Exercises - Page 152: 60


amplitude = $1$ period = $120^o$ or $\frac{2}{3}\pi$

Work Step by Step

The given graph shows around two periods of the sine function. Since the highest point is equivalent to two squares from the x-axis, and each square representing 0.5, then the amplitude is: $=2 \times 0.5 \\=1$ Horizontally, one period of the graph is from the y-axis (or $x=0$) up to the edge of the fourth square to the right of the y-axis. . Since each square represents $30^o$ or $\frac{\pi}{6}$, then the period of the given function is: $=30^o \times 4 \\=120^o$ or $=\frac{\pi}{6} \times 4 \\=\frac{2}{3}\pi$
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