How to Draw a Regular 17-gon 237
for k = 1, . . . , 16, so
x
1
= 2(cos θ + cos 8θ + cos 4θ + cos 2θ )
x
2
= 2(cos 3θ + cos 7θ + cos 5θ + cos 6θ )
y
1
= 2(cos θ + cos 4θ )
y
2
= 2(cos 8θ + cos 2θ ) (20.4)
y
3
= 2(cos 3θ + cos 5θ )
y
4
= 2(cos 7θ + cos 6θ )
Equation (20.1) implies that
x
1
+ x
2
= −1
Now (20.4) and the identity
2cosmθ cos nθ = cos(m + n)θ + cos(m −n)θ
imply that
x
1
x
2
= 4(x
1
+ x
2
) = −4
using (20.3). Hence x
1
and x
2
are zeros of the quadratic polynomial
t
2
+t −4 (20.5)
Further, x
1
> 0 so that x
1
> x
2
. By further trigonometric expansions,
y
1
+ y
2
= x
1
y
1
y
2
= −1
and y
1
,y
2
are the zeros of
t
2
−x
1
t −1 (20.6)
Further, y
1
> y
2
. Similarly, y
3
and y
4
are the zeros of
t
2
−x
2
t −1 (20.7)
and y
3
> y
4
. Now
2cosθ + 2 cos 4θ = y
1
4cosθ cos 4θ = 2 cos 5θ + 2 cos 3θ = y
3
so
z
1
= 2cosθ z
2
= 2cos4θ
are the ...