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Trigonometry Values Pi/3

From Trigonometry Values Pi/6

$\displaystyle \sin\left({\pi\over 6}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}$ (1)
$\displaystyle \cos\left({\pi\over 6}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\sqrt{3}$ (2)

together with the trigonometric identity
\begin{displaymath}
\sin(2\alpha) = 2\sin\alpha\cos\alpha,
\end{displaymath} (3)

the identity
\begin{displaymath}
\sin\left({\pi\over 3}\right)= 2\sin\left({\pi\over 6}\right...
...extstyle{1\over 2}}\sqrt{3}\,) ={\textstyle{1\over 2}}\sqrt{3}
\end{displaymath} (4)

is obtained. Using the identity
\begin{displaymath}
\cos(2\alpha) = 1-2\sin^2\alpha,
\end{displaymath} (5)

then gives
\begin{displaymath}
\cos\left({\pi\over 3}\right)= 1-2\sin^2\left({\pi\over 6}\right)= 1-2({\textstyle{1\over 2}})^2 = {\textstyle{1\over 2}}.
\end{displaymath} (6)

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

See also Equilateral Triangle




© 1996-9 Eric W. Weisstein
1999-05-26