Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 4 - Graphs of the Circular Functions - Section 4.1 Graphs of the Sine and Cosine Functions - 4.1 Exercises - Page 147: 59f

Answer

$16^{\circ}$F

Work Step by Step

Since December $31$ represents Day $365$, we substitute $x=365$ in the formula: $T(365)=37+21\sin[\frac{2\pi}{365}(x-91)]$ $T(365)=37+21\sin[\frac{2\pi}{365}(365-91)]$ $T(365)=37+21\sin[\frac{2\pi}{365}(274)]$ $T(365)=37+21\sin[4.717]$ $T(365)=37+21(-1.000)$ $T(365)=37-21$ $T(365)=16^{\circ}$F Therefore, on Day $365$, the temperature is $16^{\circ}$F.
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