Trigo compound angles

tan (5π/12)
= tan (π/4 + π/6)
= (tan π/4 + tan π/6) / (1 - tan π/4 tan π/6)
= (1 + 1/√3) / (1 - 1 . 1/√3)
= (√3 + 1) / (√3 - 1)

2007-02-22 08:53:00 補充：
　R sin (θ＋α)= R (sin θ cos α＋cos θ sin α)= (R cos α) sin θ＋(R sin α) cos θ.　√3 (cos θ＋sin θ)＋(cos θ－sin θ)= (√3－1) sin θ＋(√3＋1) cos θ

2007-02-22 08:53:20 補充：
So,R cos α = (√3－1) and R sin α = (√3＋1)tan α = (√3＋1) / (√3－1)　　 = tan (5π/12)For acute α, α = 5π/12.tan (5π/12) = (√3＋1) / (√3－1)cos (5π/12) = (√3－1) / 2R cos (5π/12) = √3－1R = 2=== finished ===

因為tan(5π/12)=(√3+1)/(√3-1)
所以√3+1=(√3-1)*tan(5π/12)&√3-1=(√3+1)/tan(5π/12)
√3(cosθ+sinθ)+(cosθ-sinθ)
=(√3+1)*cosθ+(√3-1)*sinθ
=(√3-1)*tan(5π/12)*cosθ+(√3-1)*sinθ
=(√3-1)*sin(5π/12)*cosθ/cos(5π/12)+(√3-1)*sinθ
=[(√3-1)/cos(5π/12)]*[sin(5π/12)*cosθ+sinθ*cos(5π/12)]
=[(√3-1)/cos(5π/12)]*[sin(5π/12+θ)]
=Rsin(θ+α)
又因為cos(5π/12)=(√2+√6)/4
即R=(√3-1)/cos(5π/12)=4(√3-1)/(√2+√6)=2√2(2-√3)
α=5π/12