One more of those

I will use these trigonometric identity repeatedly:

$\cos \left ( \frac{2\pi}{n} \right) + \cos \left ( \frac{4\pi}{n} \right) + \cos \left ( \frac{6\pi}{n} \right) + \ldots + \cos \left ( \frac{(n-1)\pi}{n} \right) = -\frac12$ for all positive odd number $n\geq 3$ .

And $\cos \left ( \frac{\pi}{n} \right) + \cos \left ( \frac{3\pi}{n} \right) + \cos \left ( \frac{5\pi}{n} \right) + \ldots + \cos \left ( \frac{(n-2)\pi}{n} \right) = -\frac12$ for all positive odd number $n\geq 3$ .

The sum becomes

$\begin{aligned} \displaystyle W &=& \sum_{\text{gcd}(k,45)=1 }^{45} \cos \left( \frac {k \pi}{45} \right) \\ &=& \displaystyle \sum_{k \text{ odd} }^{43} \cos \left( \frac {k \pi}{45} \right) + \displaystyle \sum_{k \text{ even}}^{44} \cos \left( \frac {k \pi}{45} \right) - \displaystyle \sum_{\text{gcd}(k,45)\ne 1 }^{45} \cos \left( \frac {k \pi}{45} \right) \\ &=& - \displaystyle \sum_{\text{gcd}(k,45)\ne 1 }^{45} \cos \left( \frac {k \pi}{45} \right) \\ -W &=& \displaystyle \sum_{\text{gcd}(k,45)=3 }^{45} \cos \left( \frac {k \pi}{45} \right) + \displaystyle \sum_{\text{gcd}(k,45)=5 }^{45} \cos \left( \frac {k \pi}{45} \right) - \displaystyle \sum_{\text{gcd}(k,45)=15 }^{45} \cos \left( \frac {k \pi}{45} \right) \\ &=& \displaystyle \sum_{\text{gcd}(k,15)=1 }^{15} \cos \left( \frac {k \pi}{15} \right)+ \displaystyle \sum_{\text{gcd}(k,9)=1 }^{5} \cos \left( \frac {k \pi}{9} \right) - \displaystyle \sum_{\text{gcd}(k,3)=1 }^{3} \cos \left( \frac {k \pi}{3} \right) \\ &=& \displaystyle \sum_{\text{gcd}(k,15)=1 }^{15} \cos \left( \frac {k \pi}{15} \right)+ \displaystyle \sum_{\text{gcd}(k,9)=1 }^{5} \cos \left( \frac {k \pi}{9} \right) \\ &=& -\displaystyle \sum_{k \in S_1 } \cos \left( \frac {k \pi}{15} \right) - \displaystyle \sum_{k \in S_2} \cos \left( \frac {k \pi}{15} \right) \\ \end{aligned}$

Where $S = \{ 3,5,6,9,10,12 \}, S_2 = \{3,6 \}$ .

Apply $\cos(A) = -\cos(\pi - A)$ and simplify everything, we get $W = \boxed 0$ .

I don't have a mind-blowing solution, as Pi Han hoped for... just a simple counting argument, using the principle of inclusion/exclusion.

For a divisor $d$ of $360$ , we define $S_d=\sum_{k=1}^{360/d}\cos(k\times{d^\circ})$ . Using roots of unity or symmetry, we see that $S_d=0$ for all $d<360$ . Our sum is related to $S_8$ , but with the multiples of $24^\circ$ and $40^\circ$ omitted; thus we try $S_8-S_{24}-S_{40}$ . But now we are subtracting too much; the common multiples of $24^\circ$ and $40^\circ$ are subtracted twice. Since $lcm(24,40)=120$ , we have to add $S_{120}$ to correct for this overcount ("inclusion/exclusion").Thus our sum is $S_8-S_{24}-S_{40}+S_{120}=\boxed{0}$ .

One can write a very short solution in terms of primitive roots of unity and the Möbius Function $\mu(n)$ : Our sum is the sum of the primitive $45$ -th roots of unity, which is $\mu(45)=0$ since $45$ fails to be square-free.