Find the Maclaurin polynomial of degree 5 for the function. $f ( x ) = \sin ( 2 x )$
A) $2 x - \frac { 4 } { 3 } x ^ { 3 } + \frac { 4 } { 15 } x ^ { 5 }$
B) $2 + \frac { 4 } { 3 } x ^ { 2 } + \frac { 2 } { 3 } x ^ { 2 }$
C) $2 x - \frac { 8 } { 3 } x ^ { 3 } + \frac { 32 } { 5 } x ^ { 5 }$
D) $2 x - \frac { 4 } { 3 } x ^ { 3 } + \frac { 2 } { 3 } x ^ { 5 }$
E) $2 x + \frac { 4 } { 3 } x ^ { 3 } + \frac { 4 } { 15 } x ^ { 5 }$

Question

Quizplus · Accepted Answer

The Maclaurin series for $\sin(x)$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$. For $\sin(2x)$, replace $x$ with $2x$ to get $2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!}$, which simplifies to $2x - \frac{8x^3}{6} + \frac{32x^5}{120}$, or $2x - \frac{4x^3}{3} + \frac{4x^5}{15}$.

Find the Maclaurin Polynomial of Degree 5 for the Function f(x)=sin⁡(2x)f ( x ) = \sin ( 2 x )f(x)=sin(2x)

Find the Maclaurin Polynomial of Degree 5 for the Function $f ( x ) = \sin ( 2 x )$