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Find the exact value of:

$$\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+ \ldots +\cos 358^{\circ}+\cos 359^{\circ}$$

I got $0$ as I did this by assigning either a positive or negative $x$ variable for each quadrant. Is this method valid and my answer right?

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    If you added $\cos 0$, then your analysis would be correct. Therefore, the actual value is...? – 2017-02-06
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    That's why you've to be careful when using 'intuitive reasoning'. Better to work out the algebra to verify it. In this case you compare $\cos(x)$ and $\cos(x+180^\circ)$. – 2017-02-07
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    Use http://math.stackexchange.com/questions/117114/sum-cos-when-angles-are-in-arithmetic-progression – 2017-02-07
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    Possible duplicate of [$\sum \cos$ when angles are in arithmetic progression](http://math.stackexchange.com/questions/117114/sum-cos-when-angles-are-in-arithmetic-progression) – 2017-02-07

5 Answers 5

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Almost. Observe that every term except for $\cos(180^{\circ})$ will cancel out.

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    [Simplicity is simpler than complexity.](http://meta.math.stackexchange.com/a/25800/276406) ;) – 2017-02-07
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The answer is $-1$.

My solution uses vector addition (equivalently, complex number addition).

Think of $\cos n^\circ$ as the $x$-coordinate of the point $(\cos n^\circ,\sin n^\circ)$, the point $n$ degrees along the circle. The sum: $$\sum_{n=0}^{359}(\cos n^\circ,\sin n^\circ)$$ is, therefore, the sum of the vertices of a regular $360$-gon centered at the origin. (Notice that it begins at $n=0$, not $n=1$.)

This sum doesn't change when you rotate it by $1^\circ$ around the origin, since rotating a $360$-gon by $1^\circ$ doesn't change it, and since rotation around the origin is a linear map (the rotation of the sum is the sum of the rotations). The only vector that doesn't change when you rotate it around the origin is the origin itself, $(0,0)$. Therefore, the above sum equals $(0,0)$.

Looking at the $x$-coordinate, we get: \begin{align} 0&=\cos0^\circ+\cos1^\circ+\dotsb+\cos359^\circ\\ -\cos0^\circ&= \phantom{\cos0^\circ+{}\!}\cos1^\circ+\dotsb+\cos359^\circ\\ -1&= \phantom{\cos0^\circ+{}\!}\cos1^\circ+\dotsb+\cos359^\circ\\ \end{align}

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You can use Lagrange's trigonometric identity $$\sum_{n=1}^{N}\cos(n\theta)=-\frac{1}{2}+\frac{\sin((N+\frac{1}{2})\theta)}{2\sin(\frac{\theta}{2})}$$ Substituting $N=359$ and $\theta=\frac{\pi}{180}$, you get $$-\frac{1}{2}+\frac{\sin(360-\frac{1}{2})\frac{\pi}{180}}{2\sin(\frac{1}{2}\frac{\pi}{180})}=-\frac{1}{2}+\frac{\sin(2\pi-\frac{\pi}{360})}{2\sin(\frac{\pi}{360})}=-1$$ since $\sin(2\pi-\theta)=-\sin(\theta)$.

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    @Brian Tung You are absolutely correct. I will edit the answer. (Should the comments stay, original answer used incorrect, as pointed out by Brian Tung, $\theta=\frac{\pi}{360}$.) – 2017-02-06
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    You can say more than just $\not= 0$; since $\sin(2\pi-\frac{\pi}{360}) = -\sin \frac{\pi}{360}$, we can equate it to $-1$. – 2017-02-06
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    @BrianTung I wasn't going to be so explicit, but since some of the other answers already were, I guess there is no point. – 2017-02-06
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    Perhaps. I wouldn't necessarily change it for that reason, though. – 2017-02-06
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Almost. Just write $$\sum_{k=1}^{359}\cos\left ( \frac{2k\pi}{360}\right)=\cos (\pi)+\sum_{k=1}^{179}\cos\left ( \frac{2k\pi}{360}\right)+\sum_{k=181}^{359}\cos\left ( \frac{2k\pi}{360}\right) $$ Grouping in Gauss Sums yields $$\sum_{k=1}^{179}\cos\left ( \frac{2k\pi}{360}\right)=\sum_{k=1}^{89}\cos \left (\frac{2k\pi}{360}\right)+ \cos\left (\pi-\frac{2k\pi}{360}\right)$$ $$\sum_{k=181}^{359}\cos\left ( \frac{2k\pi}{360}\right)=\sum_{k=1}^{89}\cos \left (\pi+\frac{2k\pi}{360}\right)+ \cos\left (2\pi-\frac{2k\pi}{360}\right)$$ Each term cancels out because $$\cos(\pi-\alpha)=-\cos (\alpha)$$ $$\cos(\pi+\alpha)=-\cos(\alpha)=-\cos(2\pi-\alpha)$$ A proof without words would go like this: enter image description here

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$$\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+ \ldots +\cos 358^{\circ}+\cos 359^{\circ}=$$ $$\tfrac{2\sin0.5^{\circ}\cos 1^{\circ}+2\sin0.5^{\circ}\cos 2^{\circ}+2\sin0.5^{\circ}\cos 3^{\circ}+ \ldots +2\sin0.5^{\circ}\cos 358^{\circ}+2\sin0.5^{\circ}\cos 359^{\circ}}{2\sin0.5^{\circ}}$$ and you'll get it

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    Is this a joke? – 2017-02-07
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    You can now apply one of the trigonometric [product identities](https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities) to each term in the numerator. (This is more complicated than @SarahRubenstein's method but it's more obvious what's left after cancelling terms.) – 2017-02-07