Sine^2 Sum

Let the expression by $S$ , then we have:

$\begin{aligned} S & = \sum_{k=1}^{360} \sin^2 k^\circ \\ & = \sum_{k=1}^{360} \frac{1}{2} \left(1 - \cos 2 k^\circ \right) \\ & = \frac{1}{2} \left( \sum_{k=1}^{360} 1 - \color{#3D99F6}{\sum_{k=1}^{360} \cos 2 k^\circ} \right) \quad \quad \small \color{#3D99F6}{\text{See } Note.} \\ & = \frac{1}{2} \left( 360 - \color{#3D99F6}{0} \right) \\ & = \boxed{180} \end{aligned}$

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$\color{#3D99F6}{Note:}$

We note that:

$\begin{aligned} \cos 2k^\circ & = - \cos (180-2k)^\circ \\ \Rightarrow \sum_{k=1}^{90} \cos 2k^\circ & = \cos 180^\circ = -1 \\ \sum_{k=1}^{180} \cos 2k^\circ & = \cos 180^\circ + \cos 360^\circ & = -1+1 = 0 \\ \Rightarrow \sum_{k=1}^{360} \cos 2k^\circ & = 0 \end{aligned}$

Notice that $\sin^2(x)=\sin^2(180-x)=\sin^2(x-180)$ Thus $\sum_{k=1}^{360}\sin(x)=\sum_{k=1}^{180}2\sin(x)$ Also we see that $\sin^2(x)=\cos^2(90-x)=\cos^2(x-90)$ Therefore $\sum_{k=1}^{180}2\sin(x)=\sum_{k=1}^{180}\sin^2(x)+\cos^2(x)=\sum_{k=1}^{180}1=180$ So by transitivity, we must have $\sum_{k=1}^{360}\sin(x)=\boxed{180}$

$\sin^2(x^{\circ})$ is symmetric about $y=180^{\circ}$ so

$\label{eq1} \begin{aligned} \sum_{i=1}^{360}\sin^2(i^{\circ}) &= \int_{0^{\circ}}^{360^{\circ}} \sin^2(x)dx \\ &= \int_0^{360}\sin^2(\frac{2\pi}{360}x)dx \\ & u = \frac{2\pi}{360}x \Leftrightarrow du = \frac{2\pi}{360}dx \\ &= \frac{360}{2\pi}\int_0^{360}\sin^2(x)dx \\ &= \frac{360}{2\pi} \bigg[ \frac{1}{2}(u-\sin(u)\cos(u)) \bigg]_0^{360} \\ & u = \frac{2\pi}{360}x = \begin{cases} 2\pi, u = 360 \\ 0, u = 0 \end{cases} \\ &=\frac{360}{2\pi} \bigg[ (\frac{1}{2}(2\pi-\sin(2\pi)\cos(2\pi))) - (\frac{1}{2}(0-\sin(0)\cos(0))) \bigg] \\ &=\frac{360}{4\pi} \bigg[ 2\pi - 0\cdot 1 - 0 + 0\cdot 1 \bigg] \\ &= \frac{360\cdot 2\pi}{4\pi} = \frac{360}{2} = 180 \end{aligned}$