Đáp án:
$B. S = 2018$
Giải thích các bước giải:
Ta có:
$\dfrac{1}{\sin x} = \dfrac{\sin\dfrac{x}{2}}{\sin\dfrac{x}{2}.\sin x}$
$= \dfrac{\sin\left(x -\dfrac{x}{2}\right)}{\sin\dfrac{x}{2}.\sin x}$
$=\dfrac{\sin x.\cos\dfrac{x}{2} - \sin\dfrac{x}{2}.\cos x}{\sin\dfrac{x}{2}.\sin x}$
$= \dfrac{\sin x.\cos \dfrac{x}{2}}{\sin\dfrac{x}{2}.\sin x} - \dfrac{\sin\dfrac{x}{2}.\cos x}{\sin\dfrac{x}{2}.\sin x}$
$= \cot\dfrac{x}{2} - \cot x$
Do đó:
$\dfrac{1}{\sin x} = \cot\dfrac{x}{2} - \cot x$
Tương tự, ta được:
$\dfrac{1}{\sin2x} = \cot x- \cot2x$
$\dots$
$\dfrac{1}{\sin2^{2018}x} = \cot2^{2017}x -\cot2^{2018}x$
Ta được:
$\mathop{\sum}\limits_{k=0}^{2018}\dfrac{1}{\sin2^kx} = 0$
$\Leftrightarrow \cot\dfrac{x}{2} - \cot2^{2018}x = 0$
$\Leftrightarrow \cot2^{2018}x = \cot\dfrac{x}{2}$
$\Leftrightarrow 2^{2018}x = \dfrac{x}{2} + k\pi$
$\Leftrightarrow 2^{2019}x = x + k2\pi$
$\Leftrightarrow (2^{2019} - 1)x = k2\pi$
$\Leftrightarrow x = \dfrac{k2\pi}{2^{2019} -1}$
$\Rightarrow \begin{cases}a = 2019\\b = 1\end{cases}$
$\Rightarrow S = a - b = 2019 - 1 = 2018$
