Given a Right Triangle with angles defined to be
and
, it must be true that
![\begin{displaymath}
\alpha+2\alpha+{\textstyle{1\over 2}}\pi = \pi,
\end{displaymath}](t_1906.gif) |
(1) |
so
. Define the hypotenuse to have length
and the side opposite
to have length
, then the
side opposite
has length
. This gives
and
![\begin{displaymath}
\sin(2\alpha) = \sqrt{1-x^2}.
\end{displaymath}](t_1911.gif) |
(2) |
But
![\begin{displaymath}
\sin(2\alpha)=2\sin\alpha\cos\alpha=2x\sqrt{1-x^2},
\end{displaymath}](t_1912.gif) |
(3) |
so we have
![\begin{displaymath}
\sqrt{1-x^2}=2x\sqrt{1-x^2}\,.
\end{displaymath}](t_1913.gif) |
(4) |
This gives
, or
![\begin{displaymath}
\sin\left({\pi\over 6}\right)= {\textstyle{1\over 2}}.
\end{displaymath}](t_1915.gif) |
(5) |
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}](t_1917.gif) |
(6) |
Summarizing,
See also Hexagon, Hexagram
© 1996-9 Eric W. Weisstein
1999-05-26