Precalculus, Chapter 5, 5.4, Section 5.4, Problem 14

You need to find the values of sine and cosine of -pi/12 , using the formulas sin(a-b) = sin a*cos b - sin b*cos a and cos(a-b) = cos a*cos b + sin a*sin b , such that:
sin(-pi/12) = sin(pi/6 - pi/4) = sin(pi/6)cos(pi/4) - sin(pi/4)cos(pi/6)
sin(-pi/12) = 1/2*sqrt2/2 - sqrt2/2*sqrt3/2
sin(-pi/12) =sqrt2/2*(1-sqrt3)/2
cos(-pi/12) = cos(pi/6 - pi/4) = cos(pi/6)cos(pi/4) + sin(pi/4)sin(pi/6)
cos(-pi/12) = sqrt3/2*sqrt2/2 + sqrt2/2*1/2
cos(-pi/12) = sqrt2/2*(1+sqrt3)/2
You need to evaluate tangent function such that:
tan(-pi/12) = (sin(-pi/12) )/(cos(-pi/12) )
tan(-pi/12) = (1-sqrt3)/(1+sqrt3)
tan(-pi/12) = -((1-sqrt3)^2)/2
Hence, evaluating the values of the functions yields sin(-pi/12) =sqrt2/2*(1-sqrt3)/2, cos(-pi/12) = sqrt2/2*(1+sqrt3)/2, tan(-pi/12) = -((1-sqrt3)^2)/2.