Trigonometry (11th Edition) Clone

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

Chapter 5 - Trigonometric Identities - Section 5.1 Fundamental Identities - 5.1 Exercises - Page 200: 27



Work Step by Step

We take a look at the point where $x=2\pi$ At $x=2\pi$, $y=1$ Then we take a look at the point where $x=-2\pi$. At $x=-2\pi$, $y=1$ Therefore, $f(-2\pi)=f(2\pi)=1$ We can take another case where $x=\pi$ and $x=-\pi$. We find that $f(\pi)=f(-\pi)=-1$ In fact, as we take a close examination, the graph has an $Oy$ symmetry, which means every 2 points where $x=a$ and $x=-a$ has the same value of $y=f(a)$. In other words, $f(-x)=f(x)$
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