Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.4 Taylor Polynomials - Exercises - Page 493: 24

Answer

\begin{aligned} \sin 3 \theta =3 \theta-9 \theta^{3}+\ldots \ldots+\frac{(-1)^{n} 3^{n}}{(2 n+1) !} \theta^{2 n+1} \end{aligned}

Work Step by Step

Since \begin{aligned} f(\theta) &=\sin 3 \theta & & f(0)=0 \\ f^{\prime}(\theta) &=3 \cos 3 \theta & & f^{\prime}(0) =3 \\ f^{\prime \prime}(\theta) &=-9 \sin 3 \theta & & f^{\prime \prime}(0) =0 \\ f^{\prime \prime \prime}(\theta) &=-27 \cos 3 \theta & f^{\prime \prime \prime}(0) &=-27 \\ f^{(4)}(\theta) &=81 \sin 3 \theta & f^{(4)}(0) &=0 \end{aligned}Then \begin{aligned} \sin 3 \theta &=f(0)+\frac{f^{\prime}(0)}{1 !} \theta+\frac{f^{\prime \prime}(0)}{2 !} \theta^{2}+\frac{f^{\prime \prime \prime}(0)}{3 !} \theta^{3}+\ldots \ldots+\frac{f^{(n)}(0)}{n !} \theta^{n} \\ &=3 \theta-9 \theta^{3}+\ldots \ldots+\frac{(-1)^{n} 3^{n}}{(2 n+1) !} \theta^{2 n+1} \end{aligned}
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