Tangent of 22.5 Degrees/Proof 2

Theorem

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

Proof

\(\ds \tan 22.5 \degrees\)

\(=\)

\(\ds \dfrac {\sin 22.5 \degrees} {\cos 22.5 \degrees}\)

\(\ds \)

\(=\)

\(\ds \dfrac {\dfrac 1 2 \sqrt {2 - \sqrt 2} } {\dfrac 1 2 \sqrt {2 + \sqrt 2} }\)

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

$\blacksquare$