Radical Expressions For Trigonometric Functions

[nycmath]

This exercise shows how to obtain radical expressions for sin 18° and cos 18°, using the figure.

A. Let x = common lengths AC, BC, and BD.
Use similar triangles to deluce that
[tex]\frac{x}{1+x}[/tex] = 1/x. Then show that
x = [tex]\frac{1+ \sqrt{5} }{2}[/tex]

B. In [tex]\triangle[/tex]BDC, draw an altitude from B to segment DC, meeting segment CD at F. Use right triangle BFC to conclude that sin 18° = [tex]\frac{1}{1+ \sqrt{5} }[/tex].

C. Rationalize the denominator in part B to obtain
sin 18° = [tex]\frac{ \sqrt{5} -1}{4}[/tex].