Answer
$0$
Work Step by Step
We know that $\sin{(\theta)}=\cos{(90^{\circ}-\theta)}$, hence $\sin{(40^{\circ})}-\cos{(50^{\circ})}=\cos{(90^{\circ}-40^{\circ})}-\cos{(50^{\circ})}=\cos{(50^{\circ})}-\cos{(50^{\circ})}=0.$
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 - Chapter Test - Page 559: 2
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