作等腰直角三角形\(\Delta{ABC}\)，\(\overline{AC}=\overline{BC}\)，\({D}\)為\(\overline{BC}\)，上一點，使得\(\angle{BCD}={15}^{\circ} \)﹒
設\(\overline{AD}={2}\)﹒則\(\overline{CD}=1\)，\(\overline{AC}=\sqrt{3}\)，\(\overline{BC}=\sqrt{3}-1 \)﹒
作\(\overline{DE} \perp \overline{AB}\)於\({E}\)，因為\(\Delta{BED}\)等腰直角三角形，所以\(\overline{DE}=\frac{\sqrt{3}-1}{\sqrt{2}}\)﹒
故\({\rm sin}{15}^\circ =\frac{\frac{\sqrt{3}-1}{\sqrt{2}}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\)﹒
