Answer
$\frac{3-3\sqrt2}{2}$.
Work Step by Step
$2\sin^2{60^\circ}-3\cos{45^\circ}=2\left(\frac{\sqrt3}{2}\right)^2-3\frac{\sqrt2}{2}=2\frac{3}{4}-\frac{3\sqrt2}{2}=\frac{3}{2}-\frac{3\sqrt2}{2}=\frac{3-3\sqrt2}{2}$.
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 6 - Trigonometric Functions - Chapter Review - Chapter Test - Page 436: 12
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