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

Question

Quizplus · Accepted Answer

The Maclaurin series for $\cos(x)$ is $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n } } { ( 2 n ) ! }$. For $\cos(-8x)$, replace $x$ with $-8x$, resulting in $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( - 8 x ) ^ { 2 n } } { ( 2 n ) ! }$, which simplifies to $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 8 ^ { 2 n } x ^ { 2 n } } { ( 2 n ) ! }$ because $(-8)^{2n} = 8^{2n}$.

Find the Maclaurin Series for the Given Function cos⁡(−8x)\cos ( - 8 x )cos(−8x)

Find the Maclaurin Series for the Given Function $\cos ( - 8 x )$