Precalculus, Chapter 5, 5.4, Section 5.4, Problem 20

sin((-7pi)/12)=-sin((7pi)/12)
=-sin(pi/3+pi/4)
using the identity sin(x+y)=sin(x)cos(y)+cos(x)sin(y)
=-(sin(pi/3)cos(pi/4)+cos(pi/3)sin(pi/4))
=-(sqrt(3)/2*1/sqrt(2)+1/2*1/sqrt(2))
=-(sqrt(3)+1)/(2sqrt(2))
rationalizing the denominator,
=(-sqrt(2)(sqrt(3)+1))/4
cos((-7pi)/12)=cos((7pi)/12)
=cos(pi/3+pi/4)
using the identity cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
=cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4)
=(1/2*1/sqrt(2)-sqrt(3)/2*1/sqrt(2))
=(1-sqrt(3))/(2sqrt(2))
rationalizing the denominator,
=(sqrt(2)(1-sqrt(3)))/4
=(sqrt(2)-sqrt(6))/4
tan((-7pi)/12)
=sin((-7pi)/12)/cos((-7pi)/12)
plug in the values evaluated above,
=((-sqrt(2)(sqrt(3)+1))/4)/((sqrt(2)-sqrt(6))/4)
=(-sqrt(2)(sqrt(3)+1))/(sqrt(2)-sqrt(6))
rationalize the denominator,
=-((sqrt(6)+sqrt(2))(sqrt(2)+sqrt(6)))/((sqrt(2)-sqrt(6))(sqrt(2)+sqrt(6)))
=-(2sqrt(3)+6+2+2sqrt(3))/(2-6)
=-(4sqrt(3)+8)/(-4)
=sqrt(3)+2