Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.3 Sum and Difference Identities - 7.3 Exercises: 101

Answer

$\frac{\sin(s-t)}{\sin t}+\frac{\cos(s-t)}{\cos t}=\frac{\sin s}{\sin t\cos t}$

Work Step by Step

Start with the left side: $\frac{\sin(s-t)}{\sin t}+\frac{\cos(s-t)}{\cos t}$ Expand using the sum and difference formulas for sine and cosine: $=\frac{\sin s\cos t-\cos s\sin t}{\sin t}+\frac{\cos s\cos t+\sin s\sin t}{\cos t}$ Get a common denominator: $=\frac{(\sin s\cos t-\cos s\sin t)\cos t}{\sin t\cos t}+\frac{(\cos s\cos t+\sin s\sin t)\sin t}{\sin t\cos t}$ $=\frac{\sin s\cos^2 t-\cos s\sin t\cos t}{\sin t\cos t}+\frac{\cos s\cos t\sin t+\sin s\sin^2 t}{\sin t\cos t}$ $=\frac{\sin s\cos^2 t-\cos s\sin t\cos t+\cos s\cos t\sin t+\sin s\sin^2 t}{\sin t\cos t}$ Note that the second and third terms in the numerator cancel. Simplify: $=\frac{\sin s\cos^2 t-\cos s\sin t\cos t+\cos s\sin t\cos t+\sin s\sin^2 t}{\sin t\cos t}$ $=\frac{\sin s\cos^2 t+\sin s\sin^2 t}{\sin t\cos t}$ $=\frac{\sin s(\cos^2+\sin^2 t)}{\sin t\cos t}$ $=\frac{\sin s*1}{\sin t\cos t}$ $=\frac{\sin s}{\sin t\cos t}$ Since this equals the right side, the identity has been proven.
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