Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 5 - Section 5.1 - Proving Identities - 5.1 Problem Set - Page 280: 69


Counterexample: $\displaystyle \theta=-\frac{\pi}{2}$ (sample answer)

Work Step by Step

The problem that leads us to search for a counterexample is that on the RHS, we have a POSITIVE square root of a number. So, we find a $\theta$ for which $\sin\theta$ is negative. Sine is negative in quadrants III and IV, so we can select $ \displaystyle \theta=-\frac{\pi}{2}$ (sample) LHS$=-1$ RHS=$\sqrt{1-0}=+1$ LHS$\neq$RHS so the identity is not valid.
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