Start with an equilateral triangle with each side of length 1. The line from any vertex perpendicular to the opposite side divides that triangle into two right triangles. Further, the that line bisects the angle at the vertex **and** bisects the opposite side. Since all three angles in an equilateral triangle are $\pi/3$ radians (60 degrees) the angle at the bisected vertex is $\pi/6$ (30 degrees).

Further, the length of the side "adjacent" to the $\pi/6$ (30 degrees) angle is $\frac{1}{2}$ and, by the Pythagorean theorem, the side "opposite" the $\pi/3$ (60 degrees) angle is $\sqrt{1- 1/4}= \sqrt{3/4}= \frac{\sqrt{3}}{2}$ while the hypotenuse has length 1.

So $sin(\pi/3)= sin(60)= \frac{\frac{\sqrt{3}}{2}}{1}= \frac{\sqrt{3}}{2}$

$cos(\pi/3)= \cos(60)= \frac{\frac{1}{2}}{1}= \frac{1}{2}$

$tan(\pi/3)= tan(60)= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}= \sqrt{3}$.

secant, cosecant, and cotangent are the reciprocals of those.

And, of course, the complementary angle, $\pi/6$ (30 degrees) has

$cos(\pi/3)= sin(60)= \frac{\frac{\sqrt{3}}{2}}{1}= \frac{\sqrt{3}}{2}$

$sin(\pi/3)= \cos(60)= \frac{\frac{1}{2}}{1}= \frac{1}{2}$

$cot(\pi/3)= tan(60)= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}= \sqrt{3}$.

Another easy angle is $\pi/4$ radians (45 degrees). Since that is half a right angle, an isosceles right triangle with legs of length 1 has both acute angles of measure $\pi/4$ radians (45 degrees) and, by the Pythagorean theorem, hypotenuse of length

$\sqrt{2}$.

So $sin(\pi/4)= sin(45)= cos(\pi/4)= cos(45)= \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}$

$tan(\pi/4)= tan(45)= cot(\pi/4)= cot(45)= \frac{1}{1}= 1$.