The Taylor series representation of $f ( x ) = \sin x$ at $\frac { \pi } { 4 }$ is

A) $\sum _ { n = 0 } ^ { \infty } \frac { \sin \left( \frac { \pi ( 2 n + 1 ) } { 4 } \right) \left( x - \frac { \pi } { 4 } \right) ^ { n } } { n ! }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { \sin \left( \frac { \pi ( 2 n + 3 ) } { 4 } \right) \left( x - \frac { \pi } { 4 } \right) ^ { n } } { n ! }$
C) $\sum _ { n = 0 } ^ { \infty } \frac { \sin \left( \frac { \pi ( 2 n - 1 ) } { 4 } \right) \left( x - \frac { \pi } { 4 } \right) ^ { n } } { n ! }$
D) $\sum _ { n = 0 } ^ { \infty } \frac { \sin \left( \frac { \pi ( 2 n - 3 ) } { 4 } \right) \left( x - \frac { \pi } { 4 } \right) ^ { n } } { n ! }$
E) $\sum _ { n = 0 } ^ { \infty } \frac { \sin \left( \frac { \pi ( 2 n + 5 ) } { 4 } \right) \left( x - \frac { \pi } { 4 } \right) ^ { n } } { n ! }$

Question

The Taylor series representation of $f ( x ) = \sin x$ at $\frac { \pi } { 4 }$ is

A) $\sum _ { n = 0 } ^ { \infty } \frac { \sin \left( \frac { \pi ( 2 n + 1 ) } { 4 } \right) \left( x - \frac { \pi } { 4 } \right) ^ { n } } { n ! }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { \sin \left( \frac { \pi ( 2 n + 3 ) } { 4 } \right) \left( x - \frac { \pi } { 4 } \right) ^ { n } } { n ! }$
C) $\sum _ { n = 0 } ^ { \infty } \frac { \sin \left( \frac { \pi ( 2 n - 1 ) } { 4 } \right) \left( x - \frac { \pi } { 4 } \right) ^ { n } } { n ! }$
D) $\sum _ { n = 0 } ^ { \infty } \frac { \sin \left( \frac { \pi ( 2 n - 3 ) } { 4 } \right) \left( x - \frac { \pi } { 4 } \right) ^ { n } } { n ! }$
E) $\sum _ { n = 0 } ^ { \infty } \frac { \sin \left( \frac { \pi ( 2 n + 5 ) } { 4 } \right) \left( x - \frac { \pi } { 4 } \right) ^ { n } } { n ! }$

Quizplus · Accepted Answer

The Taylor series of $f(x) = \sin x$ about $x = a$ is given by $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n$, where $f^{(n)}(a)$ is the $n$th derivative of $f$ evaluated at $a$. For $\sin x$ at $x = \frac{\pi}{4}$, the derivatives alternate between $\sin$ and $\cos$, both of which are $\frac{\sqrt{2}}{2}$ at $x = \frac{\pi}{4}$. The correct pattern for the derivatives of $\sin x$ at $\frac{\pi}{4}$ matches the formula in option A, considering the sine function's periodicity and derivative sequence.

The Taylor Series Representation Of f(x)=sin⁡xf ( x ) = \sin xf(x)=sinx

The Taylor Series Representation Of $f ( x ) = \sin x$