info prev up next book cdrom email home

Trigonometry Values Pi/6

Given a Right Triangle with angles defined to be $\alpha$ and $2\alpha$, it must be true that

\begin{displaymath}
\alpha+2\alpha+{\textstyle{1\over 2}}\pi = \pi,
\end{displaymath} (1)

so $\alpha = {\pi/6}$. Define the hypotenuse to have length $1$ and the side opposite $\alpha$ to have length $x$, then the side opposite $2\alpha$ has length $\sqrt{1-x^2}$. This gives $\sin\alpha\equiv x$ and
\begin{displaymath}
\sin(2\alpha) = \sqrt{1-x^2}.
\end{displaymath} (2)

But
\begin{displaymath}
\sin(2\alpha)=2\sin\alpha\cos\alpha=2x\sqrt{1-x^2},
\end{displaymath} (3)

so we have
\begin{displaymath}
\sqrt{1-x^2}=2x\sqrt{1-x^2}\,.
\end{displaymath} (4)

This gives $2x=1$, or
\begin{displaymath}
\sin\left({\pi\over 6}\right)= {\textstyle{1\over 2}}.
\end{displaymath} (5)

$\cos(\pi/6)$ is then computed from
\begin{displaymath}
\cos\left({\pi\over 6}\right)= \sqrt{1 - \sin^2\left({\pi \o...
...{\textstyle{1\over 2}})^2} = {\textstyle{1\over 2}}\sqrt{3}\,.
\end{displaymath} (6)

Summarizing,
$\displaystyle \sin\left({\pi\over 6}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}$ (7)
$\displaystyle \cos\left({\pi\over 6}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\sqrt{3}$ (8)
$\displaystyle \tan\left({\pi\over 6}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 3}}\sqrt{3}\,.$ (9)

See also Hexagon, Hexagram




© 1996-9 Eric W. Weisstein
1999-05-26