Tangent of 67.5 Degrees
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Theorem
- $\tan 67.5 \degrees = \tan \dfrac {3 \pi} 8 = \sqrt 2 + 1$
where $\tan$ denotes tangent.
Proof
| \(\ds \tan 67.5 \degrees\) | \(=\) | \(\ds \map \tan {45 \degrees + 22.5 \degrees}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\tan 45 \degrees + \tan 22.5 \degrees} {1 - \tan 45 \degrees \tan 22.5 \degrees}\) | Tangent of Sum | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {1 + \paren {\sqrt 2 - 1} } {1 - 1 \times \paren {\sqrt 2 - 1} }\) | Tangent of $45 \degrees$ and Tangent of $22.5 \degrees$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\sqrt 2} {2 - \sqrt 2}\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\sqrt 2 \paren {2 + \sqrt 2} } {\paren {2 - \sqrt 2} \paren {2 + \sqrt 2} }\) | multiplying top and bottom by $2 + \sqrt 2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 \sqrt 2 + 2} {4 - 2}\) | Difference of Two Squares | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sqrt 2 + 1\) | simplifying |
$\blacksquare$