[SOLVED] What is cos(pi/9)?

We are looking for the real solution to $\displaystyle \begin{align*} 4x^3 - 3x = \frac{1}{2} \end{align*}$ . A brief look at the graph shows that there are, in fact, three real solutions to this equation.

If we try to solve $\displaystyle \begin{align*} 4x^3 - 3x = \frac{1}{2} \end{align*}$ using the Cubic Formula, first note that this is the same as $\displaystyle \begin{align*} x^3 - \frac{3}{4}x = \frac{1}{8} \end{align*}$ , a depressed cubic (i.e., one without an $\displaystyle \begin{align*} x^2 \end{align*}$ term).

Now we look for two numbers, s and t, so that $\displaystyle \begin{align*} 3st = -\frac{3}{4} \end{align*}$ and $\displaystyle \begin{align*} \end{align*}s^3 - t^3 = \frac{1}{8}$ . It turns out that $\displaystyle \begin{align*} x = s - t \end{align*}$ will be a solution to our cubic equation.

So now we need to find s and t. First note that $\displaystyle \begin{align*} s = -\frac{1}{4t} \end{align*}$ , and substituting gives

$\displaystyle \begin{align*} \left( -\frac{1}{4t} \right)^3 - t^3 &= \frac{1}{8} \\ -\frac{1}{64t^3} - t^3 &= \frac{1}{8} \\ -\frac{1}{64} - t^6 &= \frac{t^3}{8} \\ 1 + 64t^6 &= -8t^3 \\ 64t^6 + 8t^3 + 1 &= 0 \\ 64T^2 + 8T + 1 &= 0 \textrm{ if we let } T = t^3 \\ T &= \frac{8 \pm \sqrt{8^2 - 4(64)(1)}}{2(64)} \\ T &= \frac{8 \pm \sqrt{64 - 256}}{128} \\ T &= \frac{ 8 \pm \sqrt{ -192 } }{128} \\ T &= \frac{ 8 \pm 8\sqrt{ 3 }\, i }{128} \\ T &= \frac{1 \pm \sqrt{3}\,i}{16} \\ t^3 &= \frac{1 \pm \sqrt{3}\,i}{16} \\ t &= \sqrt[3]{\frac{1 \pm \sqrt{3}\,i}{16}} \end{align*}$

Just taking the first solution, $\displaystyle \begin{align*} t = \sqrt[3]{\frac{1 + \sqrt{3}\,i}{16}} = \frac{\sqrt[3]{1 + \sqrt{3}\,i}}{2\sqrt[3]{2}} \end{align*}$ and remembering that

$\displaystyle \begin{align*} s &= -\frac{1}{4t} \\ &= -\frac{1}{4 \left( \frac{\sqrt[3]{1 + \sqrt{3}\,i}}{2\sqrt[3]{2}} \right) } \\ &= -\frac{\sqrt[3]{2}}{2\sqrt[3]{1 + \sqrt{3}\,i}} \end{align*}$

which means one of the solutions to the cubic is

$\displaystyle \begin{align*} x &= s - t \\ &= -\frac{\sqrt[3]{2}}{2\sqrt[3]{1 + \sqrt{3}\,i}} - \frac{\sqrt[3]{1 + \sqrt{3}\,i}}{2\sqrt[3]{2}} \end{align*}$

Surprisingly, this is a real number, so you will be able to long divide your cubic to find a resulting quadratic equation, and then use the Quadratic Formula to find the remaining two roots. One of these three roots will be $\displaystyle \begin{align*} \cos{\left( \frac{\pi}{9} \right)} \end{align*}$ . Alternatively, you could use the other solution $\displaystyle \begin{align*} t = \sqrt[3]{\frac{1 - \sqrt{3}\,i}{16}} \end{align*}$ to get a second root.

PHEW!