Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 2 - Review Exercises - Page 93: 11


$\cos A = \sin (90^{\circ} -A) = \sin B$

Work Step by Step

Use cofunctions $\cos x = \sin (90^{\circ} -x)$ In a triangle, you have : $m(\angle B) = 90^{\circ} - m(\angle A)$ Therefore using the cofunction formula above $\cos A = \sin (90^{\circ} -A) = \sin B$
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