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Trigonometry Values Pi/6

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

\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
\sin(2\alpha) = \sqrt{1-x^2}.
\end{displaymath} (2)

\end{displaymath} (3)

so we have
\end{displaymath} (4)

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

$\cos(\pi/6)$ is then computed from
\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)

$\displaystyle \sin\left({\pi\over 6}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}$ (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