Answer
$0$
Work Step by Step
We know that $\cos(\theta)=\sin(90^{\circ}-\theta)$.
Hence,
$\cos(62^{\circ})-\sin(28^{\circ})\\
=\sin(90^{\circ}-62^{\circ})-\sin(28^{\circ})\\
=\sin(28^{\circ})-\sin(28^{\circ})\\
=0$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 8 - Applications of Trigonometric Functions - Chapter Review - Review Exercises - Page 557: 3
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