It's kinda silly, but trig as taught revolves mostly around two right triangles: the 30,60,90 and the 45,45,90. Learn those and you'll do well in trig.
The 30,60,90 right triangle is actually half an equilateral triangle. Let #s# be the side length of an equilateral triangle and #h# be the height. The height splits the equilateral triangle into two right triangles. The hypotenuse of each is the side #s#, the short side #s/2# (since it's half the equilateral triangle's side), and the long side is the height #h#. By the Pythagorean Theorem
#s^2 = (s/2)^2 + h^2#
#s^2 = s^2/4 + h^2#
#h^2 = \frac 3 4 s^2#
# h = \frac {sqrt{3}}{2} s#
We can set our hypotenuse #s=2# giving sides # 1 # and # \sqrt{3} #.
#\pi/3# or #60^\circ# is the angle opposite the long side. Let's calculate all its trig functions:
# cos frac \pi 3 = cos 60^\circ = frac{ text{ adjacent } }{ text{ hypotenuse } } = 1/2 #
# sin frac \pi 3 = sin 60^\circ = frac{ text{ opposite } }{ text{ hypotenuse } } = sqrt{3}/2 #
# tan \frac pi 3 = tan 60^\circ = frac{ text{ opposite } }{ text{ adjacent } } = sqrt{3} #
The remaining trig functions are the reciprocals:
# sec frac \pi 3 = sec 60^\circ = frac{ text{ hypotenuse } }{ text{ adjacent } } = 2 #
# csc frac \pi 3 = csc 60^\circ = frac{ text{ hypotenuse } }{ text{ opposite } } = 2/sqrt{3} = 2 / 3 sqrt{3} #
# cot \frac pi 3 = cot 60^\circ = frac{ text{ adjacent } }{ text{ opposite } } = 1/sqrt{3} = \sqrt{3}/3 #
