Answer
See proof
Work Step by Step
Write
$\mathop \sum \limits_{k = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{{2^k}}}{\sin ^{\left( {k + 1} \right)}}x = \sin x - \dfrac{1}{2}{\sin ^2}x + \dfrac{1}{4}{\sin ^3}x - \dfrac{1}{8}{\sin ^4}x + \cdot\cdot\cdot$
$\mathop \sum \limits_{k = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{{2^k}}}{\sin ^{\left( {k + 1} \right)}}x = \mathop \sum \limits_{k = 0}^\infty \left( {\sin x} \right){\left( { - \dfrac{{\sin x}}{2}} \right)^k}$
The series on the right-hand side converges when it is a geometric series with $a = \sin x$ and $\left| r \right| = \left| { - \dfrac{{\sin x}}{2}} \right| \lt 1$. That is,
$\left| {\dfrac{{\sin x}}{2}} \right| \lt 1$, ${\ \ \ \ \ \ \ }$ $\left| {\sin x} \right| \lt 2$
Since $ - 1 \le \sin x \le 1$, the inequality above is always true. Therefore, it converges for all real values of $x$. And the sum by 9.3.3 Theorem is
$\mathop \sum \limits_{k = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{{2^k}}}{\sin ^{\left( {k + 1} \right)}}x = \mathop \sum \limits_{k = 0}^\infty \left( {\sin x} \right){\left( { - \dfrac{{\sin x}}{2}} \right)^k} = \dfrac{{\sin x}}{{1 - \left( { - \dfrac{{\sin x}}{2}} \right)}} = \dfrac{{2\sin x}}{{2 + \sin x}}$, ${\ \ \ }$ for all real values of $x$
Hence,
$\sin x - \dfrac{1}{2}{\sin ^2}x + \dfrac{1}{4}{\sin ^3}x - \dfrac{1}{8}{\sin ^4}x + \cdot\cdot\cdot = \dfrac{{2\sin x}}{{2 + \sin x}}$
