The proof is simple. If f : R→R,  f(x) = cos (ax) + cos x would have period p, then f(p) = f(0), so, cos (ap)=1,  cos (p)=1.
This implies ap = 2mπ, p = 2nπ, (m,n ∈ Z), whence a = m/n ∈ Q, contradiction, q.e.d.
I don't think a graphical proof exists.

Here is a really counterintuitive fact: the identity function  x ↦ x on R can be written as a sum of two periodic functions!

Edit: corrected π/2 + 2mπ  by  2mπ and π/2 + 2nπ  by  2nπ.

It will be enough to choose a set of 10 points as below.

restart
#    1
#   2 3
#  4 5 6
# 7 8 9 0
tri := 123,245,253,356,478,485,589,596,690,  146,279,380,  286,493,  170:
tri:=eval(convert~([tri],base, 10),0=10); # the triangles; one of them should be monochromatic
# tri := [[3, 2, 1], [5, 4, 2], [3, 5, 2], [6, 5, 3], [8, 7, 4], 
#   [5, 8, 4], [9, 8, 5], [6, 9, 5], [10, 9, 6], [6, 4, 1], [9, 7, 2], [10, 8, 3], [6, 8, 2], [3, 9, 4], [10, 7, 1]]

#tri:=remove(has, tri, 1); #### remove vertex 1 for testing ####
P:=combinat:-permute([0$10,1$10],10):nops(P): #all possible colorings (brute force)
for p in P do
  ok := false;
  for t in tri do
    if p[t[1]]=p[t[2]] and p[t[1]]=p[t[3]] then ok:=true; break fi
  od;
  if not ok then print("color_counterex"=p); break fi;
od:
if ok then print("OK!") fi;

                             "OK!"

# When testing (uncomment tri:= remove(...)) ==> "color_counterex" = [0,0,0,1,1,1,1,0,0,1]
 

latex(``* e, 'output'='string')

P.S. In the worksheet the code is better formatted.
Download ABC-vv.mw

n must be >3
The polygon with maximal area is the cyclic one.
Denote L[i] the sides, here L[i] = i, and r the radius of the circle.
Let a[i] be the angle A[i]OA[i+1], where A[1], ..., A[n] are the vertices
and O is the center of the circumscribed circle. Then 

Sum(arcsin(L[i]/r/2, i=1..n) = Pi  ==> r

restart
R := n -> local i,r; fsolve(sum(arcsin(i/r/2), i=1..n) - Pi, r =n/2 .. 2*n):

Vector( 17, k -> 'R'(k+3)=R(k+3));

n:=7:  # Graphic example
r:=R(n):
b[0]:=0:
for i to n do a[i] := 2*arcsin(i/r/2); b[i]:=b[i-1]+a[i] od:
MaxArea:=r^2/2*add(sin(a[i]), i=1..n); 
plot([seq]([cos(b[i]),sin(b[i])], i=0..n), axes=none);

        MaxArea := 54.72494665

evala(%);
#           2

Hint for a student solution "by hand":
-  Use  (a+b)^3 = a^3 + b^3 + 3*a*b*(a + b)

-- Obtain a cubic equation satisfied by your number x and factor it as (x-2)(...)=0. 

1/2*evalf(Int(min(-6*cos(t), 2-2*cos(t))^2, t=Pi/2..3*Pi/2));
identify(%);
#                          15.70796327
#                              5 Pi

Using Edge is ok, Firefox too; maybe try some font settings:

The answer for your previous problem works here too.

restart;
eq:=x^2+y^2-N*(1+x*y):  # [x=0, y=sqrt(N)] and [x=sqrt(N),y=0] are solutions!
XY:=[X=x, Y=2*y-x*N]: xy:=solve(XY,[x,y])[]:
EQ:=simplify(eval(eq,xy)):
for N in [9, 49 , 729] do
  SOL:=isolve(EQ):
  sol[N]:=map(u -> simplify(eval(xy, u)), [SOL]);
  num:={seq}(simplify(sol[N])[],_Z1=0..2);
  print('N'=N, select(type,num, [anything=integer,anything=integer]));
od:

N = 9, {[x = -1497363, y = -13307787], 
  [x = -1497363, y = -168480], [x = -168480, y = -1497363], 
  [x = -168480, y = -18957], [x = -18957, y = -168480], 
  [x = -18957, y = -2133], [x = -2133, y = -18957], 
  [x = -2133, y = -240], [x = -240, y = -2133], 
  [x = -240, y = -27], [x = -27, y = -240], [x = -27, y = -3], 
  [x = -3, y = -27], [x = -3, y = 0], [x = 0, y = -3], 
  [x = 0, y = 3], [x = 3, y = 0], [x = 3, y = 27], 
  [x = 27, y = 3], [x = 27, y = 240], [x = 240, y = 27], 
  [x = 240, y = 2133], [x = 2133, y = 240], 
  [x = 2133, y = 18957], [x = 18957, y = 2133], 
  [x = 18957, y = 168480], [x = 168480, y = 18957], 
  [x = 168480, y = 1497363], [x = 1497363, y = 168480], 
  [x = 1497363, y = 13307787]}

N = 49, {[x = -96687343207, y = -4735705783543], 
  [x = -96687343207, y = -1974033600], 
  [x = -1974033600, y = -96687343207], 
  [x = -1974033600, y = -40303193], 
  [x = -40303193, y = -1974033600], [x = -40303193, y = -822857], 
  [x = -822857, y = -40303193], [x = -822857, y = -16800], 
  [x = -16800, y = -822857], [x = -16800, y = -343], 
  [x = -343, y = -16800], [x = -343, y = -7], [x = -7, y = -343], 
  [x = -7, y = 0], [x = 0, y = -7], [x = 0, y = 7], 
  [x = 7, y = 0], [x = 7, y = 343], [x = 343, y = 7], 
  [x = 343, y = 16800], [x = 16800, y = 343], 
  [x = 16800, y = 822857], [x = 822857, y = 16800], 
  [x = 822857, y = 40303193], [x = 40303193, y = 822857], 
  [x = 40303193, y = 1974033600], [x = 1974033600, y = 40303193], 
  [x = 1974033600, y = 96687343207], 
  [x = 96687343207, y = 1974033600], 
  [x = 96687343207, y = 4735705783543]}

N = 729, {[x = -4052517025117644747, y = -2954279352292037818803], 
  [x = -4052517025117644747, y = -5559018725201760], 
  [x = -5559018725201760, y = -4052517025117644747], 
  [x = -5559018725201760, y = -7625554438293], 
  [x = -7625554438293, y = -5559018725201760], 
  [x = -7625554438293, y = -10460313837], 
  [x = -10460313837, y = -7625554438293], 
  [x = -10460313837, y = -14348880], 
  [x = -14348880, y = -10460313837], [x = -14348880, y = -19683], 
  [x = -19683, y = -14348880], [x = -19683, y = -27], 
  [x = -27, y = -19683], [x = -27, y = 0], [x = 0, y = -27], 
  [x = 0, y = 27], [x = 27, y = 0], [x = 27, y = 19683], 
  [x = 19683, y = 27], [x = 19683, y = 14348880], 
  [x = 14348880, y = 19683], [x = 14348880, y = 10460313837], 
  [x = 10460313837, y = 14348880], 
  [x = 10460313837, y = 7625554438293], 
  [x = 7625554438293, y = 10460313837], 
  [x = 7625554438293, y = 5559018725201760], 
  [x = 5559018725201760, y = 7625554438293], 
  [x = 5559018725201760, y = 4052517025117644747], 
  [x = 4052517025117644747, y = 5559018725201760], 
  [x = 4052517025117644747, y = 2954279352292037818803]}

With sol[N] you may inspect the general solution depending on _Z1.

Download dioph-sols-vv.mw

select(u -> nops(u)=m and andmap(isprime,u), L);