Quartic Sins!

Note that $50 = 60 - 10$ and $70 = 60 - 10$ . Apply the compound angle formula: $\sin(A\pm B) = \sin(A) \cos(B) \pm \cos(A) \sin(B)$ :

$\sin(50^\circ) = \frac{ \sqrt3}2 \cos(10^\circ) + \frac12 \sin(10^\circ) \\ \sin(70^\circ) = \frac{ \sqrt3}2 \cos(10^\circ) - \frac12 \sin(10^\circ)$

By binomial expansion

$\begin{aligned} \sin^4(50^\circ) + \sin^4(70^\circ) &=& \frac18 [9\cos^4 (10^\circ) + 18\cos^2(10^\circ)\sin^2 (10^\circ) + \sin^4 (10^\circ) ] \\ \sin^4(10^\circ) + \sin^4(50^\circ) + \sin^4(70^\circ) &=& \frac18 [9\cos^4 (10^\circ) + 18\cos^2(10^\circ)\sin^2 (10^\circ) + 9\sin^4 (10^\circ)] \\ &=& \frac98 (\cos^2(10^\circ) + \sin^2(10^\circ) )^2 = \boxed{\frac98} \end{aligned}$

$\sin^4{10^\circ} + \sin^4{50^\circ} + \sin^4{70^\circ} \\ = \left( \dfrac{1 - \cos{20^\circ}}{2} \right)^2 + \left( \dfrac{1 - \cos{100^\circ}}{2} \right)^2 + \left( \dfrac{1 - \cos{140^\circ}}{2} \right)^2 \\ = \frac{1}{4} [(1 + \cos{160^\circ} )^2 + (1 + \cos{80^\circ} )^2 +(1 + \cos{40^\circ})^2] \\ = \frac{1}{4} [1 + 2\cos{40^\circ} + \color{#D61F06}{\cos^2{40^ \circ}} + 1 + 2\cos{80^\circ} + \color{#D61F06}{\cos^2{80^\circ}} \\ \quad \quad + 1 + 2\cos{160^\circ} + \color{#D61F06}{\cos^2{160^\circ}}] \\ = \frac{1}{4} \left[3 + 2(\color{#3D99F6}{\cos{40^\circ} + \cos{80^\circ} + \cos{120^\circ}} - \cos{120^\circ} + \color{#3D99F6}{\cos{160^\circ}} ) \\ \quad \quad +\color{#D61F06}{ \frac{1}{2}(\cos{80^\circ} + 1 + \cos{160^\circ} + 1 + \cos{320^\circ} + 1)} \right] \\ = \frac{1}{4} \left[3 \color{#3D99F6}{ - \frac{1}{2}} + \frac{1}{2} + \frac{1}{2}(3 + \color{#D61F06}{\cos{40^\circ}} + \cos{80^\circ} + \cos{120^\circ} - \cos{120^\circ} \cos{160^\circ}) \right] \\ = \frac{1}{4} \left[3 + \frac{1}{2} \left(3 - \frac{1}{2} + \frac{1}{2} \right) \right] = \frac{1}{4} \left[3 + \frac{3}{2} \right] = \frac{9}{8} = \boxed{1.125}$

$sin^{4}10^{\circ}+sin^{4}50^{\circ}+sin^{4}70^{\circ}$

$=(sin^{2}10^{\circ})^{2}+(sin^{2}50^{\circ})^{2}+(sin^{2}70^{\circ})^{2}$

$=(\frac{1-cos20^{\circ}}{2})^{2}+(\frac{1-cos100^{\circ}}{2})^{2}+(\frac{1-cos140^{\circ}}{2})^{2}$

$=\frac{1}{4}\cdot \left ( 1-2cos20^{\circ}+cos^{2}20^{\circ}+1-2cos100^{\circ}+cos^{2}100^{\circ}+1-2cos140^{\circ}+cos^{2}140^{\circ} \right )$

$=\frac{1}{4}\cdot \left [ 3-2\left ( cos20^{\circ}+cos100^{\circ}+cos140^{\circ} \right )+\left ( cos^{2}20^{\circ}+cos^{2}100^{\circ}+cos^{2}140^{\circ} \right ) \right ]$

$\because \: \: cos20^{\circ}+cos100^{\circ}+cos140^{\circ}$

$\: \: \: \: =cos(60^{\circ}-40^{\circ})+cos(60^{\circ}+40^{\circ})+cos140^{\circ}$

$\: \: \: \: 2\: cos60^{\circ}cos40^{\circ}+cos140^{\circ}=cos40^{\circ}+cos140^{\circ}=0$

Also $\because \: \: cos^{2}20^{\circ}+cos^{2}100^{\circ}+cos^{2}140^{\circ}$

$\: \: \: \: =\frac{cos40^{\circ}+1}{2}+\frac{cos200+1^{\circ}}{2}+\frac{cos280^{\circ}+1}{2}$

$\: \: \: \: =\frac{1}{2}\left ( 3+cos40^{\circ}+cos200^{\circ}+cos280^{\circ} \right )$ ,

and $cos40^{\circ}+cos200^{\circ}+cos280^{\circ}$

$\: \: =cos(120^{\circ}-80^{\circ})+cos(120^{\circ}+80^{\circ})+cos280^{\circ}$

$\: \: =2 \: cos120^{\circ}cos80^{\circ}+cos280^{\circ}=-cos80^{\circ}+cos280^{\circ}=0$

$\therefore cos^{2}20^{\circ}+cos^{2}100^{\circ}+cos^{2}140^{\circ}=\frac{1}{2}\left ( 3+cos40^{\circ}+cos200^{\circ}+cos280^{\circ} \right )$

$=\frac{3}{2}$

$\therefore sin^{4}10^{\circ}+sin^{4}50^{\circ}+sin^{4}70^{\circ}=\frac{1}{4}\cdot (3+\frac{3}{2})= \boxed{1.125}$