University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 28: 37


If $A=B$, then $$\cos(A-B)=\cos A\cos B+\sin A\sin B=1$$ It agrees with the already known identity $\cos^2A+\sin^2A=1$

Work Step by Step

$$\cos(A-B)=\cos A\cos B+\sin A\sin B$$ Since $A=B$, we replace $B$ with $A$, then we have: - The left side: $\cos(A-B)=\cos(A-A)=\cos0=1$ - The right side: $$\cos A\cos B+\sin A\sin B=\cos A\cos A+\sin A\sin A=\cos^2A+\sin^2A=1$$ (Recall the identity $\cos^2A+\sin^2A=1$) Therefore, if $A=B$, then both sides would equal $1$. It agrees with the already known identity $\cos^2A+\sin^2A=1$
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