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 5.2 Verifying Trigonometric Identities - 5.2 Exercises - Page 210: 97


$$\sin(\csc t)=1$$ Take $t=\frac{\pi}{2}$, which meaks $\sin(\csc t)\ne1$. Therefore, the equation cannot be an identity.

Work Step by Step

$$\sin(\csc t)=1$$ To show an equation to not be an identity, we only need to find one example in which 2 sides are unequal. Here we can take $t=\frac{\pi}{2}$, but remember that there are other choices as well. $$\csc t=\csc\frac{\pi}{2}=1$$ Therefore, $$\sin(\csc t)=\sin1\approx0.841\ne1$$ (remember that $1$ here is in radian) Thus at $t=1$, 2 sides are not equal. The equation, therefore, cannot be an identity.
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