Question 577615
{{{390^o=360^o+30^o}}}, so {{{sin(390^o)=sin(30^o)}}}.
{{{sin(30^o)=1/2}}}.
I remember that, but I can also prove it out of a right triangle:
Suppose you have a right triangle ABC, with the right angle at C and a {{{30^o)}}} angle at A. That triangle is half of an equilateral triangle.
{{{drawing(300,300,-2,10,-6,6,
red(triangle(0,0,8.66,0,8.66,5)),
green(line(0,0,8.66,-5)),green(line(8.66,-5,8.66,0)),
locate(-0.5,0.35,A), locate(8.5,5.5,B), locate(8.7,0.35,C)
)}}} so {{{sin(30^o)=sin(BAC)=BC/AB=1/2}}}