Tangent of 22.5°

Problem

Compute $\tan(22{,}5°)$ using the bisection formulae for $45°$.

Solution

Since $22{,}5° = \dfrac{45°}{2}$, we use the form $\tan\!\left(\dfrac{\alpha}{2}\right) = \dfrac{1-\cos\alpha}{\sin\alpha}$ with $\alpha = 45°$, recalling that $\cos 45° = \sin 45° = \dfrac{\sqrt{2}}{2}$: $\tan(22{,}5°) = \frac{1-\cos 45°}{\sin 45°} = \frac{1-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2-\sqrt{2}}{\sqrt{2}} = \frac{2}{\sqrt{2}} - 1 = \boxed{\sqrt{2}-1}.$

Links

Topics: Trigonometric formulae
Concepts: Bisection formulae
Functions: Tangent
Methods: Bisection formulae
Skills: Calculating · Using formulae
Exercise types: Computing expressions