Precalculus, Chapter 5, 5.4, Section 5.4, Problem 25

-165^@=-(120^@+45^@)

sin(-165)=-sin(165)
sin(u+v)=sin(u)cos(v)+cos(u)sin(v)
[sin(-(u+v))]=-[sin(u)cos(v)+cos(u)sin(v)]
[sin(-(120+45))]=-[sin(120)cos(45)+cos(120)sin(45)]
[sin(-(120+45))]=-[(sqrt3/2)(sqrt2/2)+(-1/2)(sqrt2/2)]=-sqrt2/4(sqrt3+1)

cos(-165)=cos(165)
cos(u+v)=cos(u)cos(v)-sin(u)sin(v)
cos(-(u+v))=cos(u)cos(v)-sin(u)sin(v)
cos(-(120+45))=cos(120)cos(45)-sin(120)sin(45)
cos(-(120+45))=(-1/2)(sqrt2/2)-(sqrt3/2)(sqrt2/2)=-sqrt2/4(1+sqrt3)

tan(-165)=-tan(165)
tan(u+v)=(tan(u)+tan(v))/(1-tan(u)tan(v))
-tan(u+v)=-[(tan(u)+tan(v))/(1-tan(u)tan(v))]
-tan(120+45)=-[(tan(120)+tan(45))/(1-tan(120)tan(45))]=-[(-sqrt3+1)/(1-(-sqrt3)(1))]=(sqrt3-1)/(1+sqrt3)
After rationalizing the denominator the answer is 2-sqrt3.