\begin{align*} &sin^410^{\circ} +\sin^450^{\circ}+\sin^470^\circ\\ =\quad&\Big(\frac{1-\cos 20^\circ}{2}\Big)^2+\Big(\frac{1-\cos 100^\circ}{2}\Big)^2+\Big(\frac{1-\cos 140^\circ}{2}\Big)^2\\ =\quad&\frac{3}{4} -\frac{1}{2}\cdot(\cos 20^\circ + \cos 100^\circ + \cos 140^\circ) + \frac{1}{4}\cdot(\cos^2 20^\circ + \cos^2 100^\circ + \cos^2 140^\circ)\\ =\quad&\frac{3}{4} -\frac{1}{2}\cdot(2\cos 60^\circ\cos40^\circ - \cos 40^\circ)+\frac{1}{4}\Big(\frac{1+\cos 40^\circ}{2}+\frac{1+\cos 200^\circ}{2}+\frac{1+\cos 280^\circ}{2}\Big)\\ =\quad&\frac{3}{4}-0+\frac{3}{8}+\frac{1}{8}\cdot(\cos 40^\circ -\cos 20^\circ +\cos 80^\circ)\\ =\quad&\frac{9}{8} + \frac{1}{8}\cdot(-2\sin 30^\circ\sin 10^\circ+\sin 10^\circ)\\ =\quad&\boxed{\frac{9}{8}} \end{align*}