A geometry problem by Chandrashekar Giridharan

Relevant wiki: Proving Trigonometric Identities - Basic

$\begin{aligned} S & = \cos \theta^\circ + \cos (\theta^\circ + 1^\circ) + \cos (\theta^\circ + 2^\circ) + \cdots + \cos (\theta^\circ + 359^\circ) \\ & = \sum_{k=0}^{359} \cos (\theta^\circ + k^\circ) \\ & = \sum_{k=0}^{179} \cos (\theta^\circ + k^\circ) \color{#3D99F6} + \sum_{k=180}^{359} \cos (\theta^\circ + k^\circ) & \small \color{#3D99F6} \text{Note that }\cos (180^\circ - x) = - \cos x \\ & = \sum_{k=0}^{179} \cos (\theta^\circ + k^\circ) \color{#3D99F6} - \sum_{k=180}^{359} \cos (180^\circ - (\theta^\circ + k^\circ)) \\ & = \sum_{k=0}^{179} \cos (\theta^\circ + k^\circ) - \sum_{k=0}^{179} \color{#3D99F6} \cos (- \theta^\circ - k^\circ) & \small \color{#3D99F6} \text{Note that }\cos (- x) = \cos x \\ & = \sum_{k=0}^{179} \cos (\theta^\circ + k^\circ) - \sum_{k=0}^{179} \color{#3D99F6} \cos (\theta^\circ + k^\circ) \\ & = \boxed{0} \end{aligned}$

Relevant wiki: Trigonometric Periodicity Identities

The cosine value is the signed distance from the point on the unit circle to the $y$ -axis (or if written as a coordinate point, the $x$ part of the coordinate).

Each of the angle values has a matching one reflected over the $y$ -axis where the cosines are opposites. For example, $\cos(60^\circ) = -\cos(120^\circ) .$ (The only exceptions are at 90 and 270 degrees, which are both 0 to begin with.) When these pairs are added up, they equal 0. Since all angle values from 0 to 359 can be matched in this way (this represents 180 pairs, 360 angles total) the overall sum must be 0.

Relevant wiki: Sine and Cosine Graphs

The cos graph completes one cycle in 360 degrees. In those 360 values of cos, if the graph passes through some positive value x, it will always pass through negative x (-x) as well(refer the graph). Their sum cancels all the positive and negative values leaving behind an answer of 0 .

Whatever $\theta^{0}$ a.k.a starting point is, the expression above will always map a complete circle. The values of $\cos(\theta)$ from the right side of the circle will cancel out the values on the left side. Therefore, answer is $0$ .

Using Euler's formula we have: $S = \Re( \sum_{k=0}^{359} e^{i(\theta+k)} )$

$\sum_{k=0}^{359} e^{i(\theta+k)} = e^{i\theta} \sum_{k=0}^{359} e^{ik} =$

$e^{i\theta} \sum_{k=0}^{359} (e^i)^k = e^{i\theta} \sum_{k=0}^{359} a^k$ for $a = e^i$

From the formula $\sum_{k=0}^{k=n} a^k = \frac{a^{n+1}-1}{a-1}$ we have:

$S = e^{i\theta} \frac{(e^i)^{360}-1}{e^i-1} = e^{i\theta} * 0 = 0$ where the last equality comes again from Euler's formula

I hav very easy solution. You are increasing the angle, but 'cos' will have value between 1 and -1 . Now you know that sum of value of cos from angle 0° to 360° =0 . And after every 360° same cycle will continue so answer will be 0 .

The sum is zero, independent of $\theta$ being integer or not:

$\begin{aligned} \sum_{i=0}^{359} cos(\theta^\circ+i^\circ)&=\sum_{i=0}^{179} \cos(\theta^\circ+i^\circ)+\sum_{i=180}^{359}\cos(\theta^\circ+i^\circ)\\&=\sum_{i=0}^{179} \cos(\theta^\circ+i^\circ)+\sum_{i=0}^{179} \cos(\theta^\circ+i^\circ+180^\circ)\\&=\sum_{i=0}^{179} (\cos(\theta^\circ+i^\circ)-\cos(\theta^\circ+i^\circ))\\&=\sum_{i=0}^{179} 0\\&=0 \end{aligned}$

From the problem's figure of a trigonometric circle, it can be easily seen that the cosine function of an angle rotated half circle ( $180^\circ$ ) is the opposite value to the cosine of the original angle (i. e., $\cos(\theta^\circ+180^\circ)=-\cos(\theta^\circ)$ ). If we pair the first term with the 181th one, the second with 182th,... the 180th with the 359th, we get pairs of cosines of angles $180^\circ$ apart, which sums to zero. 360 angles, 180 pairs of opposed values, 180 zeros, the total sum is zero. This result is valid even if $\theta$ is not an integer.

In general,

$\boxed{\cos\theta+\cos(\theta+d)+\cos(\theta+2d)+\cos(\theta+3d)+\ldots+\cos(\theta+(n-1)d)=\frac{\cos\left(\theta+\frac{(n-1)d}{2}\right)\sin\left(\frac{nd}{2}\right)}{\sin\left(\frac{d}{2}\right)}}$

$\therefore \cos{\theta}^\circ+\cos(\theta^\circ+1^\circ)+\cos(\theta^\circ+2^\circ)+\cos(\theta^\circ+3^\circ)+\ldots+\cos(\theta^\circ+359^\circ)=\frac{\cos\left(\theta^\circ+\frac{(360-1)1}{2}\right)^\circ\sin\left(\frac{360^\circ}{2}\right)}{\sin\left(\frac{1^\circ}{2}\right)}$

$=\frac{\cos(\theta^\circ+179^\circ)\sin180^\circ}{\sin0.5^\circ}=0$

Note that $\cos(\theta + 180) = -\cos(\theta)$

We can arrange the terms of the sum such that $\cos(\theta +x)$ is paired with $\cos(\theta + x + 180)$ , for $x = 0$ to $179$ . The sum of each pair is zero, thus the entire sum is zero.

Note that this identity doesn't require $\theta$ to be an integer, or even rational.

In Turtle Geometry, it is well-known that the turtle program $\mbox{turn } \theta; \mbox{ repeat } N \mbox{ times : forward } 1; \mbox{ turn } 360^\circ/N$ traces out a regular $N$ -gon, where the turtle ends again at its starting position. The projection on the x-axis of the vertices accumulates $\cos \theta+i 360^\circ/N$ . The image illustrates this for $N=4$ , with $\phi=360^\circ/4$ and $x_k = \sum_{i=1}^{k}\cos \theta+i 360^\circ/4$ . The question involves a regular 360-gon, for which $x_{360}=0$ .

Since all cos values repeat after 360 degrees...we essentially need to calculate cos 0 + cos 1 + cos 2 + cos 3 + .....+ cos 360.

cos theta = opp/hyp for a unit circle.... = y values/1 so essentially we have to add up all the y co-ordinates at theta = 1,2,3,4,.....360 this has to cancel each other out and add up to zero