Provide an appropriate response.
-Obtain the first nonzero term of the Maclaurin series for $\sin ( \tan x ) - \tan ( \sin x )$.
A) $\frac { x ^ { 7 } } { 30 }$
B) $- \frac { x ^ { 7 } } { 30 }$
C) $\frac { x ^ { 7 } } { 60 }$
D) $- \frac { x ^ { 7 } } { 60 }$

Question

Quizplus · Accepted Answer

The Maclaurin series for sin(x) and tan(x) are:

$\sin x = x - \dfrac{x^{3}}{3!} + \dfrac{x^{5}}{5!} - \dfrac{x^{7}}{7!} + \cdots$

$	an x = x + \dfrac{x^{3}}{3} + \dfrac{2x^{5}}{15} + \dfrac{17x^{7}}{315} + \cdots$

Using these, we get:

$\sin(	an x) = \Big(	an x - \dfrac{	an^{3}x}{3!} + \dfrac{	an^{5}x}{5!} - \dfrac{	an^{7}x}{7!} + \cdots\Big)$

$	an(\sin x) = \Big(\sin x + \dfrac{\sin^{3}x}{3} + \dfrac{2\sin^{5}x}{15} + \dfrac{17\sin^{7}x}{315} + \cdots\Big)$

Therefore,

$\sin(	an x) - 	an(\sin x) = \Big(	an x - \dfrac{	an^{3}x}{3!} + \dfrac{	an^{5}x}{5!} - \dfrac{	an^{7}x}{7!} + \cdots\Big) - \Big(\sin x + \dfrac{\sin^{3}x}{3} + \dfrac{2\sin^{5}x}{15} + \dfrac{17\sin^{7}x}{315} + \cdots\Big)$

Keeping only the first nonzero term, we get:

$-\dfrac{	an^{7}x}{7!} - \dfrac{17\sin^{7}x}{315}$

Substituting $	an x = x + \dfrac{x^{3}}{3} + \dfrac{2x^{5}}{15} + \dfrac{17x^{7}}{315}$ and $\sin x = x - \dfrac{x^{3}}{3!} + \dfrac{x^{5}}{5!} - \dfrac{x^{7}}{7!} + \cdots$, we get:

$-\dfrac{(x + \frac{x^{3}}{3} + \frac{2x^{5}}{15} + \frac{17x^{7}}{315})^{7}}{7!} -  \dfrac{17(x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!})^{7}}{315}$

Expanding each term and keeping only the terms with $x^{7}$, we get:

$-\dfrac{1}{30}x^{7}$

Therefore, the first nonzero term is $\boxed{	extbf{(B) } -\dfrac{x^{7}}{30}}$

Provide an Appropriate Response sin⁡(tan⁡x)−tan⁡(sin⁡x)\sin ( \tan x ) - \tan ( \sin x )sin(tanx)−tan(sinx)

Provide an Appropriate Response $\sin ( \tan x ) - \tan ( \sin x )$