Precalculus, Chapter 5, 5.4, Section 5.4, Problem 12

You need to evaluate the sine of (7pi)/12 , using the formula sin(a+b) = sin a*cos b + sin b*cos a such that:
sin ((7pi)/12)= sin(pi/3 + pi/4) = sin (pi/3)*cos (pi/4) + sin (pi/4)*cos (pi/3)
sin ((7pi)/12)=(sqrt3)/2*(sqrt2)/2 + (sqrt2)/2*1/2
sin ((7pi)/12) = (sqrt2)/2*(sqrt3 + 1)/2
You need to evaluate the cosine of (7pi)/12 , using the formula cos(a+b) = cos a*cos b - sin b*sin a such that:
cos ((7pi)/12) = cos (pi/3 + pi/4) = cos (pi/3)*cos (pi/4)- sin ( pi/4)*sin (pi/3)
cos ((7pi)/12)= 1/2*(sqrt2)/2 - (sqrt2)/2*(sqrt3)/2
cos((7pi)/12) = (sqrt2)/2*(1 - sqrt3)/2
You need to evaluate the tangent of (7pi)/12 , such that:
tan ((7pi)/12) = (sin((7pi)/12))/(cos ((7pi)/12))
tan ((7pi)/12) = ((sqrt2)/2*(sqrt3 + 1)/2)/((sqrt2)/2*(1 - sqrt3)/2)
tan((7pi)/12) = (sqrt3 + 1)/(1 - sqrt3)
tan((7pi)/12) = ((sqrt3 + 1)*(1 + sqrt3))/(1 - 3)
tan((7pi)/12)) = -((sqrt3 + 1)^2)/2
Hence, evaluating the sine, cosine and tangent of tan(7pi)/12 , yields sin((7pi)/12 ) = (sqrt2)/2*(sqrt3 + 1)/2, cos ( (7pi)/12 ) = (sqrt2)/2*(1 - sqrt3)/2, tan (7pi)/12 = -((sqrt3 + 1)^2)/2.