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 5.4 Sum and Difference Identities for Sine and Tangent - 5.4 Exercises - Page 227: 60

Answer

When we sketch the graph of these two functions, we can see that the graphs are the same. This is an identity: $tan(\frac{\pi}{2}-\theta) = cot~\theta$
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Work Step by Step

$y = tan(\frac{\pi}{2}-\theta)$ $y = cot~\theta$ When we sketch the graph of these two functions, we can see that the graphs are the same. We can use the following identities: $sin~(-a) = -sin~a$ $cos~(-b) = cos~b$ We can demonstrate the identity: $tan(\frac{\pi}{2}-\theta) = \frac{sin(\frac{\pi}{2}-\theta)}{cos(\frac{\pi}{2}-\theta)}$ $tan(\frac{\pi}{2}-\theta) = \frac{-sin(\theta-\frac{\pi}{2})}{cos(\theta-\frac{\pi}{2})}$ $tan(\frac{\pi}{2}-\theta) = \frac{cos~\theta}{sin~\theta}$ $tan(\frac{\pi}{2}-\theta) = cot~\theta$ This is an identity: $tan(\frac{\pi}{2}-\theta) = cot~\theta$
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