Another sum of sines

There are $18$ terms. The last one is $1.$ The middle one, $\sin^2(45^\circ),$ is $0.5.$ Let $S$ be the sum of the remaining $16$ terms. Then $\begin{aligned} S &= \left(\sin^2(5^\circ) + \sin^2(85^\circ)\right) + \left(\sin^2(10^\circ) + \sin^2(80^\circ)\right) + \cdots + \left(\sin^2(40^\circ) + \sin^2(50^\circ)\right) \\ &= \left(\sin^2(5^\circ) + \cos^2(5^\circ)\right) + \left(\sin^2(10^\circ) + \cos^2(10^\circ)\right) + \cdots + \left(\sin^2(40^\circ) + \cos^2(40^\circ)\right) \\ &= 1 + 1 + \cdots + 1 = 8. \end{aligned}$ So the answer is $1 + 0.5 + 8 = \fbox{9.5}.$

$\begin{aligned} S & = \sin^2 5^\circ + \sin^2 10^\circ + \sin^2 15^\circ + \cdots + \sin^2 80^\circ + \sin^2 85^\circ + \sin^2 90^\circ \\ & = (\sin^2 5^\circ + \sin^2 85^\circ) + (\sin^2 10^\circ + \sin^2 80^\circ) + (\sin^2 15^\circ + \sin^2 75^\circ) + \cdots + (\sin^2 40^\circ + \sin^2 50^\circ) + \sin^2 45^\circ + \sin^2 90^\circ \\ & = (\sin^2 5^\circ + \cos^2 5^\circ) + (\sin^2 10^\circ + \cos^2 10^\circ) + (\sin^2 15^\circ + \cos^2 15^\circ) + \cdots + (\sin^2 40^\circ + \cos^2 40^\circ) + \sin^2 45^\circ + \sin^2 90^\circ \\ & = 1+1+1+1+1+1+1+1+\frac 12 + 1 \\ & = \boxed{9.5} \end{aligned}$