Answer
$-2$
Work Step by Step
We know that $\sin(\theta)^2+\cos(\theta)^2=1$.
Thus,
$\sin(14^{o})^2+\cos(14^{o})^2-3
\\=1-3
\\=-2$
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 - Cumulative Review - Page 437: 9
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