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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?

Ng Chung Tak
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Sarah Rubenstein
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5 Answers5

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

Nick D.
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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}

Akiva Weinberger
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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 $\sin(2\pi-\theta)=-\sin(\theta)$.

Jan
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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, $\theta=\frac{\pi}{360}$.) – Jan Feb 06 '17 at 22:21
  • You can say more than just $\not= 0$; since $\sin(2\pi-\frac{\pi}{360}) = -\sin \frac{\pi}{360}$, we can equate it to $-1$. – Brian Tung Feb 06 '17 at 23:30
  • @BrianTung I wasn't going to be so explicit, but since some of the other answers already were, I guess there is no point. – Jan Feb 06 '17 at 23:40
  • Perhaps. I wouldn't necessarily change it for that reason, though. – Brian Tung Feb 06 '17 at 23:46
1

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

Lonidard
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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

Michael Rozenberg
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