Question 1103453
<br>
{{{sin(pi/12) = (sqrt(6)-sqrt(2))/4}}}
{{{sin(17pi/12) = -sin(5pi/12) = -cos(pi/12) = -((sqrt(6)+sqrt(2))/4)}}}<br>
Note the conversions there use
{{{sin(x-pi) = -sin(x)}}}  and
{{{sin(x) = cos((pi/2)-x)}}}
So<br>
{{{sin(pi/12) - sin(17pi/12) = (sqrt(6)-sqrt(2))/4 + (sqrt(6)+sqrt(2))/4 = sqrt(6)/2}}}