Precalculus, Chapter 5, 5.4, Section 5.4, Problem 21

sin((-13pi)/12)
using the property sin(-x)=-sin(x),
sin((-13pi)/12)=-sin((13pi)/12)
=-sin(pi/2+pi/3+pi/4)
Now using sin(pi/2+x)=cos(x),
=-cos(pi/3+pi/4)
=-(cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4))
=-(1/2*1/sqrt(2)-sqrt(3)/2*1/sqrt(2))
=(sqrt(3)-1)/(2sqrt(2))
rationalizing the denominator,
=(sqrt(2)(sqrt(3)-1))/4
cos((-13pi)/12)
using the property cos(-x)=cos(x),
cos((-13pi)/12)=cos((13pi)/12)
cos((13pi)/12)=cos(pi/2+pi/3+pi/4)
now using cos(pi/2+x)=-sin(x),
=-sin(pi/3+pi/4)
=-(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
tan((-13pi)/12)=sin((-13pi)/12)/cos((-13pi)/12)
plug in the values of sin((-13pi)/12),cos((-13pi)/12) obtained above,
=((sqrt(2)(sqrt(3)-1))/4)/((-sqrt(2)(sqrt(3)+1))/4)
=-(sqrt(3)-1)/(sqrt(3)+1)
rationalizing the denominator,
=-((sqrt(3)-1)(sqrt(3)-1))/((sqrt(3)+1)(sqrt(3)-1))
=-(3+1-2sqrt(3))/2
=-(4-2sqrt(3))/2
=sqrt(3)-2