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Trigonometry Values Pi/5

Use the identity

\begin{displaymath}
\sin(5\alpha)=5\sin\alpha-20\sin^3\alpha+16\sin^5\alpha.
\end{displaymath} (1)

Now, let $\alpha\equiv{\pi/5}$ and $x\equiv\sin\alpha$. Then
\begin{displaymath}
\sin\pi=0=5x-20x^3+16x^5
\end{displaymath} (2)


\begin{displaymath}
16x^4-20x^2+5=0.
\end{displaymath} (3)

Solving the Quadratic Equation for $x^2$ gives
$\displaystyle \sin^2\left({\pi\over 5}\right)$ $\textstyle =$ $\displaystyle x^2 = {{20\pm\sqrt{(-20)^2-4\cdot 16\cdot 5}}\over 2\cdot 16}$  
  $\textstyle =$ $\displaystyle {20\pm \sqrt{80}\over 32} = {\textstyle{1\over 8}}(5\pm\sqrt{5}\,).$ (4)

Now, $\sin(\pi/5)$ must be less than
\begin{displaymath}
\sin\left({\pi\over 4}\right)= {\textstyle{1\over 2}}\sqrt{2},
\end{displaymath} (5)

so taking the Minus Sign and simplifying gives
\begin{displaymath}
\sin\left({ \pi\over 5}\right)= \sqrt{5-\sqrt{5}\over 8}={\textstyle{1\over 4}}\sqrt{10-2\sqrt{5}}.
\end{displaymath} (6)

$\cos(\pi/5)$ can be computed from
\begin{displaymath}
\cos\left({\pi\over 5}\right)= \sqrt{1-\sin^2\left({\pi\over 5}\right)} = {\textstyle{1\over 4}}(1+\sqrt{5}).
\end{displaymath} (7)

Summarizing,
$\displaystyle \sin\left({ \pi\over 5}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}\sqrt{10-2\sqrt{5}}$ (8)
$\displaystyle \sin\left({2\pi\over 5}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}\sqrt{10+2\sqrt{5}}$ (9)
$\displaystyle \sin\left({3\pi\over 5}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}\sqrt{10+2\sqrt{5}}$ (10)
$\displaystyle \sin\left({4\pi\over 5}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}\sqrt{10-2\sqrt{5}}$ (11)
$\displaystyle \cos\left({ \pi\over 5}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}(1+\sqrt{5})$ (12)
$\displaystyle \cos\left({2\pi\over 5}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}(-1+\sqrt{5}\,)$ (13)
$\displaystyle \cos\left({3\pi\over 5}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}(1-\sqrt{5}\,)$ (14)
$\displaystyle \cos\left({4\pi\over 5}\right)$ $\textstyle =$ $\displaystyle -{\textstyle{1\over 4}}(1+\sqrt{5}\,)$ (15)
$\displaystyle \tan\left({ \pi\over 5}\right)$ $\textstyle =$ $\displaystyle \sqrt{5-2\sqrt{5}}$ (16)
$\displaystyle \tan\left({2\pi\over 5}\right)$ $\textstyle =$ $\displaystyle \sqrt{5+2\sqrt{5}}$ (17)
$\displaystyle \tan\left({3\pi\over 5}\right)$ $\textstyle =$ $\displaystyle -\sqrt{5+2\sqrt{5}}$ (18)
$\displaystyle \tan\left({4\pi\over 5}\right)$ $\textstyle =$ $\displaystyle -\sqrt{5-2\sqrt{5}}\,.$ (19)

See also Dodecahedron, Icosahedron, Pentagon, Pentagram



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© 1996-9 Eric W. Weisstein
1999-05-26