Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.3 Properties of the Trigonometric Functions - 6.3 Assess Your Understanding - Page 395: 112

Answer

Odd, symmetric to the origin.

Work Step by Step

We know that $\sin$, $\csc$, $\tan$ are odd functions, which means $f(-\theta)=-f(\theta).$ We know that $\cos$, $\sec$, are even functions, which means $f(-\theta)=f(\theta).$ We know that if a graph is symmetric to the y-axis, then the points $(a,b)$ and $(-a,b)$ will be on the graph, hence $f(-b)=f(b)$, hence this is the same as the function being even. We know that if a graph is symmetric to the origin, then the points $(a,b)$ and $(-a,-b)$ will be on the graph, hence $f(-b)=-f(b)$, hence this is the same as the function being odd. We know that if a graph is symmetric to the x-axis, then the points $(a,b)$ and $(a,-b)$ will be on the graph, but then for $a$ there are two different values, hence this cannot happen for a function. We know that for the $\csc$ function, $f(-\theta)=-f(\theta)$, hence it is odd. Therefore its graph is symmetric to the origin.
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