7 Use de Moivre's theorem to show that
$$\tan 5 \theta = \frac { 5 t - 10 t ^ { 3 } + t ^ { 5 } } { 1 - 10 t ^ { 2 } + 5 t ^ { 4 } }$$
where \(t = \tan \theta\).
Deduce that the roots of the equation \(t ^ { 4 } - 10 t ^ { 2 } + 5 = 0\) are \(\pm \tan \frac { 1 } { 5 } \pi\) and \(\pm \tan \frac { 2 } { 5 } \pi\).
Hence show that \(\tan \frac { 1 } { 5 } \pi \tan \frac { 2 } { 5 } \pi = \sqrt { } 5\).
Show LaTeX source
7 Use de Moivre's theorem to show that
$$\tan 5 \theta = \frac { 5 t - 10 t ^ { 3 } + t ^ { 5 } } { 1 - 10 t ^ { 2 } + 5 t ^ { 4 } }$$
where $t = \tan \theta$.
Deduce that the roots of the equation $t ^ { 4 } - 10 t ^ { 2 } + 5 = 0$ are $\pm \tan \frac { 1 } { 5 } \pi$ and $\pm \tan \frac { 2 } { 5 } \pi$.
Hence show that $\tan \frac { 1 } { 5 } \pi \tan \frac { 2 } { 5 } \pi = \sqrt { } 5$.
\hfill \mbox{\textit{CAIE FP1 2014 Q7}}