Buktikan Tan 15°=2-akar3

Trigonometri.

[tex]\displaystyle \tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ \tan 15^{\circ}=\tan (45^{\circ}-30^{\circ})\\ =\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ}\tan 30^{\circ}}\\ =\frac{\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{3}}{1-\frac{1}{2}\sqrt{2}~\frac{1}{2}\sqrt{3}}\\ =2-\sqrt{3}[/tex]