Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Review Exercises - Page 644: 29



Work Step by Step

Since, $\sin^2 \theta +\cos^2 \theta=1$ Here, $\theta = \dfrac{\pi}{17}$ Thus, we have $\sin^2 (\dfrac{\pi}{17}) +\cos^2 (\dfrac{\pi}{17}) =1$
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