Use the identity \(\sin^2\alpha+\cos^2\alpha=1\).

\(\sin^2\alpha=1-\cos^2\alpha=1-\dfrac{21}{25}=\dfrac{4}{25}\).

So \(\sin\alpha=\pm\dfrac{2}{5}\).

Since \(\dfrac{\pi}{2}<\alpha<\pi\), the angle is in quadrant II, where \(\sin\alpha>0\).

Therefore, \(\sin\alpha=\dfrac{2}{5}\).

\(\tan 10^\circ\cdot\cot 10^\circ=1\).

\(\sin 30^\circ=\dfrac{1}{2}\).

So \(1+4\cdot\dfrac{1}{2}=1+2=3\).

\(\cos^2\alpha=1-\sin^2\alpha=1-\dfrac{15}{16}=\dfrac{1}{16}\).

So \(\cos\alpha=\pm\dfrac{1}{4}\).

Since \(-\dfrac{\pi}{2}<\alpha<0\), the angle is in quadrant IV, where \(\cos\alpha>0\).

Therefore, \(\cos\alpha=\dfrac{1}{4}\).

\(\tan 60^\circ=\sqrt{3}\), so \(\sqrt{3}\cdot\sqrt{3}=3\).

Also, \(\sin^2 20^\circ+\cos^2 20^\circ=1\).

Therefore, \(3-1=2\).

\(\sin^2\alpha=1-\cos^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\).

So \(\sin\alpha=\pm\dfrac{4}{5}\).

Since \(\pi<\alpha<\dfrac{3\pi}{2}\), the angle is in quadrant III, where \(\sin\alpha<0\).

Therefore, \(\sin\alpha=-\dfrac{4}{5}\).