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: 28



Work Step by Step

We take a look at the point where $x=\frac{\pi}{2}$ At $x=\frac{\pi}{2}$, $y=1$ Then we take a look at the point where $x=-\frac{\pi}{2}$. At $x=-\frac{\pi}{2}$, $y=-1$ Therefore, $f(-\frac{\pi}{2})=-f(\frac{\pi}{2})$ We can take another case where $x=\frac{3\pi}{2}$ and $x=-\frac{3\pi}{2}$. We find that $f(\frac{3\pi}{2})=-1$, while $f(-\frac{3\pi}{2})=1$ As we take a close examination, the graph has a symmetry through the point $O$, which means for a point where $x=a$ and $y=b$, the point where $x=-a$ would have $y=-b$. In other words, $f(-x)=-f(x)$
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