Precalculus (6th Edition) Blitzer

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

Chapter 9 - Section 9.6 - Conic Sections in Polar Coordinates - Exercise Set - Page 1032: 57

Answer

See explanations.

Work Step by Step

Step 1. $LHS=sin(2x)=2sin(x)cos(x)$ Step 2. $RHS=2cot(x)sin^2x=2(\frac{cos(x)}{sin(x)})sin^2x=2sin(x)cos(x)$ Step 3. Since $LHS=RHS$, we verified the identity.
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