I just learned that
- $\cos\left(\frac{π}{5}\right)$ = $\frac{φ}{2}$,
- $\cos\left(\frac{2π}{5}\right)$ = $\frac{1}{2φ}$ = $-\frac{ψ}{2}$,
- $\cos\left(\frac{3π}{5}\right)$ = $-\frac{1}{2φ}$ = $\frac{ψ}{2}$, and
- $\cos\left(\frac{4π}{5}\right)$ = $-\frac{φ}{2}$,
where $φ = \frac{1 + \sqrt{5}}{2}$ is the Golden Ratio and $ψ = \frac{1 – \sqrt{5}}{2}$ is the Golden Ratio Conjugate.
It seems suggestive to me that the Golden Ratio involves the square root of five and also appears in the output of the cosine function for inputs proportional to the inverse of five. So I did some digging.
The Golden Ratio is a member of the group of numbers which are solutions to the cubic equation $x^3 – a x – 1 = 0$, with $a = 1$ giving the Plastic Ratio $ρ = \sqrt[3]{\frac{9 + \sqrt{69}}{18}} + \sqrt[3]{\frac{9 – \sqrt{69}}{18}}$ and two complex roots, and $a = 2$ giving the Golden Ratio $φ$ plus the Golden Ratio Conjugate $ψ$ and $-1$. The value for $a = 3$ does not seem to have a name that I can find, but it does also show up in the cosine function.
If we denote the real-valued roots as $R_n(a)$, going from most positive at $n = 1$ to most negative at $n = 3$, then
- $\cos\left(\frac{π}{9}\right) = \frac{R_1(3)}{2} ≈ \frac{1.879385}{2}$,
- $\cos\left(\frac{2π}{9}\right) = -\frac{R_3(3)}{2} ≈ -\frac{-1.532089}{2}$,
- $\cos\left(\frac{4π}{9}\right) = -\frac{R_2(3)}{2} ≈ -\frac{-0.347296}{2}$,
- $\cos\left(\frac{5π}{9}\right) = \frac{R_2(3)}{2} ≈ \frac{-0.347296}{2}$,
- $\cos\left(\frac{7π}{9}\right) = \frac{R_3(3)}{2} ≈ \frac{-1.532089}{2}$, and
- $\cos\left(\frac{8π}{9}\right) = \frac{-R_1(3)}{2} ≈ -\frac{1.879385}{2}$.
Given that these two sets of roots for $a = 2$ and $a = 3$ show up, it makes me wonder if more do so as well. However, I haven't been able to find an way to fit the Plastic Ratio into the cosine function for $a = 1$, or the unnamed roots for $a = 4$.
Do the roots for other values of $a$ appear in the output of the cosine function? Is the appearance of these two sets of roots purely a coincidence, or is there some deeper mathematical connection?
Best Answer
Substitute $x=m \cos t$ to get $$m^3 \cos^3 t - a m \cos t -1=0$$ Let $$m=\sqrt{\frac{4a}{3}}$$ then it rearranges (assuming $a,m\ne 0$) to $$4 \cos^3 t - 3 \cos t = \sqrt{\frac{27}{4a^3}}$$ Using trigonometric identity $$\cos (3t) = \sqrt{\frac{27}{4a^3}}$$ So $$x = \sqrt{\frac{4a}{3}} \cos\left(\frac{1}{3} \cos^{-1}\left(\sqrt{\frac{27}{4a^3}}\right)\right)$$
When $a=1$ the R.H.S. of the penultimate equation is bigger than $1$ so $\cos^{-1}$ is problematic. Note that $\cos^{-1}/3$ is multi-valued.