Question 1029655
For this one we can use the identity:
{{{cos(2*(theta)) = 2*(cos(theta))^2 - 1}}}
but we'll go backwards...thus
2cos^2(7pi/12) - 1 = cos(2*(7pi/12)) = cos(7pi/6) = -cos(pi/6) =
{{{-sqrt(3) / 2}}}