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Trigonometry Values Pi/4

For a Right Isosceles Triangle, symmetry requires that the angle at each Vertex be given by

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

so $\alpha = \pi/4$. The sides are equal, so
\begin{displaymath}
\sin^2\alpha+\cos^2\alpha = 2\sin^2\alpha = 1.
\end{displaymath} (2)

Solving,
$\displaystyle \sin\left({\pi\over 4}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\sqrt{2}$ (3)
$\displaystyle \cos\left({\pi\over 4}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\sqrt{2}$ (4)
$\displaystyle \tan\left({\pi\over 4}\right)$ $\textstyle =$ $\displaystyle 1.$ (5)

See also Square




© 1996-9 Eric W. Weisstein
1999-05-26