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Find the value of $\cos 1^\circ + \cos 2^\circ + \cos 3^\circ +.....+\cos 180^\circ $

My Attempt: $$=\cos 1^\circ + \cos 2^\circ + \cos 3^\circ +....+\cos 180^\circ $$ $$= \cos 1^\circ + \cos 2^\circ + \cos 3^\circ +....-1$$

How do I complete it?

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marked as duplicate by lab bhattacharjee algebra-precalculus May 22 '17 at 1:27

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    $\begingroup$ If $x+y=180^\circ$, what can you say about $\cos(x)$ and $\cos(y)$. More specifically, can you relate $\cos(x)$ and $\cos(180^\circ - x)$ $\endgroup$ – B. Mehta May 22 '17 at 1:25
  • $\begingroup$ @labbhattacharjee, How does that question relate to mine? $\endgroup$ – pi-π May 22 '17 at 1:31
  • $\begingroup$ Just do it. What is $\cos(180-x)+\cos(x)$? $\endgroup$ – John Lou May 22 '17 at 1:33
  • $\begingroup$ @JohnLou, $\cos (180-x)+\cos x=0$. $\endgroup$ – pi-π May 22 '17 at 1:36
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    $\begingroup$ Unfortunately, I can't answer this question anymore, so I'm just gonna write it here because it seems like you need it. Your question is equal to $$\sum_{k=1}^{89}(\cos(k)) + \cos(90)+ \sum_{k=91}^{179}(\cos(k)) + \cos(180)$$ Simplifying down, we get $$\sum_{k=1}^{89}(\cos(k) + \cos(180-k)) + \cos(90) + \cos(180)$$ Notice how the summation now cancels to zero, leaving us with $$\cos(90) + \cos(180) = 0 + -1 = -1$$ $\endgroup$ – John Lou May 22 '17 at 2:12