Find the coefficient of $x ^ { 3 }$ in the Maclaurin series for $f ( x ) = \sin 2 x$ .
A) $- \frac { 2 } { 3 }$
B) $- \frac { 4 } { 3 }$
C) $\frac { 4 } { 3 }$
D) $- \frac { 8 } { 3 }$
E) $\frac { 2 } { 3 }$
F) $\frac { 8 } { 3 }$
G) $- \frac { 1 } { 3 }$
H) $\frac { 1 } { 3 }$

Question

Find the coefficient of $x ^ { 3 }$ in the Maclaurin series for $f ( x ) = \sin 2 x$ .
A) $- \frac { 2 } { 3 }$ 
B) $- \frac { 4 } { 3 }$ 
C) $\frac { 4 } { 3 }$ 
D) $- \frac { 8 } { 3 }$ 
E) $\frac { 2 } { 3 }$ 
F) $\frac { 8 } { 3 }$ 
G) $- \frac { 1 } { 3 }$ 
H) $\frac { 1 } { 3 }$

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 $2x - \frac{(2x)^3}{3!} + \cdots$, which simplifies to $2x - \frac{8x^3}{6} = 2x - \frac{4x^3}{3}$. Therefore, the coefficient of $x^3$ is $-\frac{4}{3}$.

Find the Coefficient Of x3x ^ { 3 }x3 In the Maclaurin Series For

Find the Coefficient Of $x ^ { 3 }$ In the Maclaurin Series For