解法一：cos30 °= √¯3/2
cos15 °= √¯[(1 + cos30° )/2]
sin15° = √¯[(1 - cos30° )/2]
tan15° = sin15° /cos15°
= √¯[(1 - cos30° )/(1 + cos30° )]
= √¯[(2 - √¯3)/(2 + √¯3)]
= (2 - √¯3)
解法二：作Rt△ABC,使∠A=30°∠B=90°
延长BA到D,使AD=AC,连接CD,则∠D=15°
令BG=1,则AD=AC=2,AB=√¯3.
故BD=AD+AB=2+√¯3
tan15° =tan∠D=BC/BD
=1/(2+√¯3)= (2 - √¯3)