Trigonometry 7th Edition

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

Chapter 2 - Test - Page 111: 8

Answer

$0$

Work Step by Step

$60^o$ is a special angle whose sine value is $\dfrac{\sqrt3}{2}$. $30^o$ is a special angle whose cosine value is $\dfrac{\sqrt3}{2}.$ Thus, $\sin^2{60^o} - \cos^2{30^o} \\= \left(\dfrac{\sqrt3}{2}\right)^2 - \left(\dfrac{\sqrt3}{2}\right)^2 \\=0$
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