University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 28: 40


$$\sin(2\pi-x)=-\sin x$$

Work Step by Step

$$\sin(2\pi-x)$$ Here we can use the identity that we have proved in Exercise 36: $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$ Therefore: $$\sin(2\pi-x)=\sin2\pi\cos x-\cos2\pi\sin x$$ $$\sin(2\pi-x)=0\times\cos x-1\times\sin x$$ $$\sin(2\pi-x)=-\sin x$$
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