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Trigonometry Values Pi/8


$\displaystyle \sin\left({\pi\over 8}\right)$ $\textstyle =$ $\displaystyle \sin\left({{1\over 2}\cdot{\pi\over 4}}\right)
= \sqrt{{1\over 2}\left({1-\cos{\pi\over 4}}\right)}$  
  $\textstyle =$ $\displaystyle \sqrt{{\textstyle{1\over 2}}(1-{\textstyle{1\over 2}}\sqrt{2})} = {\textstyle{1\over 2}}\sqrt{2-\sqrt{2}}.$ (1)

Now, checking to see if the Square Root can be simplified gives
\begin{displaymath}
a^2-b^2 c = 2^2-1^2\cdot 2 = 4-2 = 2,
\end{displaymath} (2)

which is not a Perfect Square, so the above expression cannot be simplified. Similarly,
$\displaystyle \cos \left({\pi \over 8}\right)$ $\textstyle =$ $\displaystyle \cos\left({{1 \over 2} {\pi \over 4}}\right)= \sqrt{{1\over 2}\left({1+\cos{\pi\over 4}}\right)}$  
  $\textstyle =$ $\displaystyle \sqrt{{1\over 2}\left({1+{\sqrt{2}\over 2}}\right)} = {\textstyle{1\over 2}}\sqrt{2+\sqrt{2}}$ (3)
$\displaystyle \tan\left({\pi \over 8}\right)$ $\textstyle =$ $\displaystyle \sqrt{{2-\sqrt{2}}\over{2+\sqrt{2}}}
= \sqrt{(2-\sqrt{2})^2 \over {4-2}} = \sqrt{{4+2-4\sqrt{2}} \over 2}$  
  $\textstyle =$ $\displaystyle \sqrt{{6-4\sqrt{2}} \over 2} = \sqrt{3-2\sqrt{2}}.$ (4)

But
\begin{displaymath}
a^2-b^2 c = 3^2-2^2 2 = 9-8 = 1
\end{displaymath} (5)

is a Perfect Square, so we can find
\begin{displaymath}
d = {\textstyle{1\over 2}}(3\pm 1) = 1, 2.
\end{displaymath} (6)

Rewrite the above as
$\displaystyle \tan\left({\pi \over 8}\right)$ $\textstyle =$ $\displaystyle \sqrt{2}-1$ (7)
$\displaystyle \cot\left({\pi\over 8}\right)$ $\textstyle =$ $\displaystyle {1 \over {\sqrt{2}-1}} = {{\sqrt{2}+1} \over {2-1}} = \sqrt{2}+1.$ (8)

Summarizing,
$\displaystyle \sin\left({\pi\over 8}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\sqrt{2-\sqrt{2}}$ (9)
$\displaystyle \sin\left({3\pi\over 8}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\sqrt{2+\sqrt{2}}$ (10)
$\displaystyle \cos \left({\pi \over 8}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\sqrt{2+\sqrt{2}}$ (11)
$\displaystyle \cos\left({3\pi\over 8}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\sqrt{2-\sqrt{2}}$ (12)
$\displaystyle \tan\left({\pi \over 8}\right)$ $\textstyle =$ $\displaystyle \sqrt{2}-1$ (13)
$\displaystyle \tan\left({3\pi\over 8}\right)$ $\textstyle =$ $\displaystyle \sqrt{2}+1.$ (14)

See also Octagon



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© 1996-9 Eric W. Weisstein
1999-05-26