Precalculus, Chapter 5, 5.4, Section 5.4, Problem 19

sin((13pi)/12)=sin(pi/2+pi/3+pi/4)
As we know that sin(pi/2+theta)=cos(theta)
:.sin((13pi)/12)=cos(pi/3+pi/4)
Now use the identity cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
sin((13pi)/12)=cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4)
sin((13pi)/12)=((1/2)(1/sqrt(2))-((sqrt(3)/2)(1/sqrt(2)))
sin((13pi)/12)=(1-sqrt(3))/(2sqrt(2))
sin((13pi)/12)=(sqrt(2)-sqrt(6))/4
cos((13pi)/12)=cos(pi/2+pi/3+pi/4)
We know that cos(pi/2+theta)=-sin(theta)
:.cos((13pi)/12)=-sin(pi/3+pi/4)
using identity sin(x+y)=sin(x)cos(y)+cos(x)sin(y)
cos((13pi)/12)=-(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))
=-(sqrt(6)+sqrt(2))/4
tan((13pi)/12)=sin((13pi)/12)/cos((13pi)/12)
plug in the values evaluated above,
tan((13pi)/12)=((sqrt(2)-sqrt(6))/4)/(-(sqrt(6)+sqrt(2))/4)
=(sqrt(2)-sqrt(6))/(-(sqrt(6)+sqrt(2)))
rationalizing the denominator,
=((sqrt(2)-sqrt(6))(sqrt(6)-sqrt(2)))/(-(6-2))
=(sqrt(12)-2-6+sqrt(12))/(-4)
=(2sqrt(12)-8)/(-4)
=(2*2sqrt(3)-8)/(-4)
=2-sqrt(3)