Chọn phương án C.

Đặt $m=\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}\cos x\cdot f(x)\mathrm{\,d}x$. Ta có $f(x)=\sin x+2m$. Khi đó $$\begin{aligned}
m&=\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}\cos x\cdot f(x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}\cos x\left(\sin x+2m\right)\mathrm{\,d}x\\
&=\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}\left(\sin x\cdot\cos x+2m\cos x\right)\mathrm{\,d}x\\ &=\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}\left(\dfrac{1}{2}\sin2x+2m\cos x\right)\mathrm{\,d}x\\
&=\left(-\dfrac{1}{4}\cos2x+2m\sin x\right)\bigg|_{0}^{\tfrac{\pi}{2}}\\
&=\dfrac{1}{2}+2m.\\
\Leftrightarrow m&=-\dfrac{1}{2}.
\end{aligned}$$
Vậy $f(x)=\sin x-1$. Suy ra $f\left(-\dfrac{\pi}{2}\right)=-1-1=-2$.