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: 23

Answer

It is unnecessary because we do not have to use Identities which involve second-power trigonometric functions.

Work Step by Step

The quadrant of $\theta$ is given to signify the signs of $\sin\theta$. In detail, if $\theta$ lies in quadrant I and II, $\sin\theta$ is positive. If $\theta$ lies in quadrant III and IV, $\sin\theta$ is negative. This fact must be given in the situation when to find $\sin\theta$, we end up with 2 results of $\sin\theta$ that are opposite in signs. These happen a lot in trigonometric identities, since a lot of identities involve dealing with the second power, which would lead to 2 opposite-signed results when we decrease to the first power. For example, look at this Pythagorean Identity, $$\sin^2\theta+\cos^2\theta=1$$ $$\sin^2\theta=1-\cos^2\theta$$ $$\sin\theta=\pm(1-\cos^2\theta)$$ As you can see, when we decrease to the first power, we end up with an opposite sign like that. Therefore, the quadrant of $\theta$ is essential to decide which result to choose. On the other hand, in exercises 17 and 18, we only need to use this identity: $$\csc\theta=\frac{1}{\sin\theta}$$ $$\sin\theta=\frac{1}{\csc\theta}$$ Here we would end up with only 1 result of $\sin\theta$. So there is no need for quadrant of $\theta$.
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