Then the triangle ABD is an isosceles triangle, ∠D+∠ABD=30, ∠D=∠ABD= 15.

Let BC = 1, then AB=AD=2 and AC=√3.

In the right triangle BCD, BC = 1, CD=2+√3, so BD=√6+√2.

Available according to the definition of trigonometric function

sin 15 =(√6-√2)/4；

cos 15 =(√6+√2)/4；

tan 15 = 2-√3；

cot 15 =2+√3。