# Mathematical proof (or disproof) please

Forgive me dopers as I have jettisoned most of my mathematical abilities, post-university, that aren’t relevant to my occupation. I am therefore not equipped to formally prove what graphically, and experimentally, seems to me to be true. And that is that for any integer ‘n’ that is not a multiple of 3, and any angle ‘a’ in degrees:

sin(a*n)+sin((a+120)*n)+sin((a+240)*n)=0.

Is this able to be arranged in a formal, step-by-step, mathematical proof or disproof?

I fell slightly embarrassed as intuitively this seems like it should be an easy one to lay out formally, but it’s past me now.

I’d do it in three parts.

1. Show that sin(120n) + sin(240n) = 0 for integers n.

2. Show that cos(120n) + cos(240n) = -1 for integers n that are not divisible by three. (It equals 2 when 3 divides n.)

3. Use sin(a + b) = sin(a)cos(b) + cos(a)sin(b) and parts 1) and 2) to manipulate your original expression into something that is identically zero when 3 divides n.

First, note that sin(a*n)+sin((a+120)*n)+sin((a+240)*n) = sin(an + 0n) + sin(an + 120n) + sin(an + 240n), and if a can be any angle, an can be any angle.

If n is an integer that is not a multiple of 3, then 0n, 120n, and 240n (in degrees, or 0n, (2/3)pin, and (4/3)pin in radians) form a set of angles that are equivalent in direction (and hence, in their sines and cosines) to 0, 120, and 240.

Then you can use sin(a + b) = sin(a)cos(b) + cos(a)sin(b) to show that
sin(an) + sin(an + 120) + sin(an + 240) = 0.

First, you can simplify the problem to showing that, for any angle a, that

sin(a) + sin(a + 120) + sin(a + 240) = 0,

because, if n is not divisible by 3, then {0, 120n, 240n} = {0, 120, 240} modulo 360, and a*n is as arbitrary an angle as a.

Now I’m tempted to end the argument by pointing out that the points on the unit circle at angles a, a+120, and a+240 are symmetric around the origin, so, considered as vectors (from the origin to these points) their sum is 0, giving the identity you seek for both cos and sin (those being their x- and y-coordinates, respectively).

If you prefer an algebraic proof, use

sin(a + 120) = sin(a)cos(120) + cos(a)sin(120) = -(1/2)sin(a) + (sqrt(3)/2)cos(a)

sin(a + 240) = sin(a)cos(240) + cos(a)sin(240) = -(1/2)sin(a) - (sqrt(3)/2)cos(a)

then add them together to see that everything cancels (and similarly for cos).

Added on preview: Ninja’d by Thudlow Boink, but he didn’t mention the geometric viewpoint and I added a little more detail.

If you like complex numbers, you can also use a neat trick that relies on the fact that rotation by an angle is the same thing as multiplication by a complex number. I’ll use radians to do this rather than degrees because it’s a little more natural in the context of complex numbers.

First note that sin(x) is the imaginary part of e[sup]ix[/sup]. So if it happens that

e[sup]ia[/sup] + e[sup]i(a + 2π/3)[/sup] + e[sup]i(a + 4π/3)[/sup]

is zero, then your original quantity is zero as well, since it’s the imaginary part of the above quantity. (The real part is the same expression with the sines replaced by cosines.)

Now use the algebraic identity that

b + b x + b x[sup]2[/sup] = b (1 - x[sup]3[/sup])/(1 - x).

If we replace b = e[sup]ia[/sup] and x = e[sup]2πi/3[/sup], then our quantity becomes

e[sup]ia[/sup] (1 - e[sup]2 π i[/sup])/(1 - e[sup]2πi/3[/sup]) = 0,

since e[sup]2πi[/sup] = 1. QED.

As a bonus, this generalizes to any set of n angles evenly spaced around the unit circle, using the identity 1 + x + … + x[sup]n-1[/sup] = (1 - x[sup]n[/sup])/(1 - x) instead.

Oh my, where did it all go?

Thanks a bunch for the replies.