Find the Maclaurin series for the given function.
-$\sin 8 x$
A) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { 2 n + 1 } 8 ^ { 2 n + 1 } x ^ { 2 n + 1 } } { n ! }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 8 ^ { 2 n + 1 } x ^ { 2 n + 1 } } { n ! }$
C) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { 2 n + 1 } 8 ^ { 2 n + 1 } x ^ { 2 n + 1 } } { ( 2 n + 1 ) ! }$
D) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 8 ^ { 2 n + 1 } x ^ { 2 n + 1 } } { ( 2 n + 1 ) ! }$

Question

Quizplus · Accepted Answer

The Maclaurin series formula for $\sin x$ is $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n + 1 } } { ( 2 n + 1 ) ! }$. In this case, the argument of $\sin$ is multiplied by a constant factor of 8, so we need to adjust the formula by replacing $x$ with $8x$. This gives us the Maclaurin series for $\sin 8x$ as $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } (8x) ^ { 2 n + 1 } } { ( 2 n + 1 ) ! }$, which matches choice D.

Find the Maclaurin Series for the Given Function sin⁡8x\sin 8 xsin8x

Find the Maclaurin Series for the Given Function $\sin 8 x$