Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Review Exercises - Page 249: 49


$sin^2~x-sin^2~y = cos^2~y-cos^2~x$

Work Step by Step

$sin^2~x-sin^2~y = (1-cos^2~x)-(1-cos^2~y)$ $sin^2~x-sin^2~y = cos^2~y-cos^2~x$
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