Tangent of 22.5 Degrees

From ProofWiki
Jump to navigation Jump to search

Theorem

$\tan 22.5 \degrees = \tan \dfrac \pi 8 = \sqrt 2 - 1$

where $\tan$ denotes tangent.


Proof 1

\(\ds \tan 22.5 \degrees\) \(=\) \(\ds \tan \dfrac {45 \degrees} 2\)
\(\ds \) \(=\) \(\ds \dfrac {1 - \cos 45\degrees} {\sin 45\degrees}\) Half Angle Formula for Tangent: Corollary $2$
\(\ds \) \(=\) \(\ds \dfrac {1 - \frac {\sqrt 2} 2} {\frac {\sqrt 2} 2}\) Cosine of $45 \degrees$ and Sine of $45 \degrees$
\(\ds \) \(=\) \(\ds \sqrt 2 - 1\) multiplying top and bottom by $\sqrt 2$

$\blacksquare$


Proof 2

\(\ds \tan 22.5 \degrees\) \(=\) \(\ds \dfrac {\sin 22.5 \degrees} {\cos 22.5 \degrees}\)
\(\ds \) \(=\) \(\ds \dfrac {\dfrac 1 2 \sqrt {2 - \sqrt 2} } {\dfrac 1 2 \sqrt {2 + \sqrt 2} }\) Sine of $22.5 \degrees$, Cosine of $22.5 \degrees$
\(\ds \) \(=\) \(\ds \dfrac {\sqrt {2 - \sqrt 2} \sqrt {2 - \sqrt 2} } {\sqrt {2 + \sqrt 2} \sqrt {2 - \sqrt 2} }\) multiplying top and bottom by $\sqrt {2 - \sqrt 2}$
\(\ds \) \(=\) \(\ds \dfrac {2 - \sqrt 2} {\sqrt {2^2 - 2} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \sqrt 2 - 1\) simplifying

$\blacksquare$


Also see