tan15° = sin15° /cos15°
= √¯[(1 - cos30° )/(1 + cos30° )]
= √¯[(2 - √¯3)/(2 + √¯3)]
= (2 - √¯3)
注:AC:BC=1:2 - √¯3
sin15° = √¯[(1 - cos30° )/2]
tan15° = sin15° /cos15°
= √¯[(1 - cos30° )/(1 + cos30° )]
= √¯[(2 - √¯3)/(2 + √¯3)]
= (2 - √¯3)
注:AC:BC=1:2 - √¯3
sin15° = √¯[(1 - cos30° )/2]