Question 786238
If you were taught that
{{{(sin(x))^2+(cos(x))^2=1}}} you would use that and the fact that {{{pi/6}}} is in the first quadrant, where sine and cosine are positive.
 
Otherwise, if you were only taught that sine and cosine are trigonmetric ratios that apply to right triangles, use a right triangle with a {{{pi/6}}} angle and hypotenuse length 1. Then invoke the Pythagorean theorem. The measures of the legs of that right triangle are {{{cos(pi/6)}}} and {{{sin(pi/6)}}}.
 
In fact, if you split an equilateral triangle in half using a median (connecting the midpoint of one side to the opposite vertex), you would get two congruent right triangles with one {{{pi/6}}} angle. Fron that idea, you can deduce both values, {{{sin(pi/6)}}} and {{{sin(pi/6)}}}. The shorter leg of those traingles, opposite the {{{pi/6}}} angle is half of the side of the equilateral triangle, that is now the right triangle's hypotenuse. Hence {{{sin(pi/6)=1/2}}}