Precalculus, Chapter 5, 5.4, Section 5.4, Problem 15

You need to evaluate the sine of 105^o , using the formula sin(a+b) = sin a*cos b + sin b*cos a such that:
sin(105^o) = sin(60^o + 45^o) = sin 60^o*cos 45^o + sin 45^o*cos 60^o
sin(105^o) = (sqrt3)/2*(sqrt2)/2 + (sqrt2)/2*1/2
sin(105^o) = (sqrt2)/2*(sqrt3 + 1)/2
You need to evaluate the cosine of 105^o , using the formula cos(a+b) = cos a*cos b - sin b*sin a such that:
cos (105^o) = cos (60^o + 45^o) = cos 60^o*cos 45^o - sin 45^o*sin 60^o
cos (105^o) = 1/2*(sqrt2)/2 - (sqrt2)/2*(sqrt3)/2
cos (105^o) = (sqrt2)/2*(1 - sqrt3)/2
You need to evaluate the tangent of 105^o , such that:
tan 105^o = (sin(105^o))/(cos (105^o))
tan 105^o = ((sqrt2)/2*(sqrt3 + 1)/2)/((sqrt2)/2*(1 - sqrt3)/2)
tan 105^o = (sqrt3 + 1)/(1 - sqrt3)
tan 105^o = ((sqrt3 + 1)*(1 + sqrt3)/(1 - 3)
tan 105^o = -((sqrt3 + 1)^2)/2
Hence, evaluating the sine, cosine and tangent of 105^o , yields sin(105^o) = (sqrt2)/2*(sqrt3 + 1)/2, cos (105^o) = (sqrt2)/2*(1 - sqrt3)/2, tan 105^o = -((sqrt3 + 1)^2)/2.