數學 M2 三角學?

(a)
BC交DE於H
DP=AP*Cosθ
DQ=BD*Cosθ
QH=BH*Cos(2π/3－θ)
RH=CH*Cos(2π/3－θ)
PQ=DQ－DP=(BD－AP)*Cosθ=xCosθ
QR=QH－RH=(BH－CH)*Cos(2π/3－θ)=xCos(2π/3－θ)
Set m=π/3－θ
2π/3－θ=m+π/3，θ=π/3－m
PR/x=PQ/x+QR/x
=Cosθ+Cos(2π/3－θ)
=Cos(π/3－m)+Cos(m+π/3)
=2Cos(π/3)Cosm
=Cosm
=Cos(π/3－θ)
Cosθ+Cos(2π/3－θ)=Cos(π/3－θ)
(b)
PR/x=PQ/x+QR/x
Cosθ+Cos(2π/3－θ)=Cosθ
(c)
Cos(π/4)+Cos(2π/3－π/4)=Cos(π/3－π/4)
Cos(π/4)+Cos(5π/12)=Cos(π/12)
Cos(π/4)+Sin(π/2-5π/12)=Cos(π/12)
Cos(π/4)+Sin(π/12=Cos(π/12)
Cos(π/12)－Sin(π/12)=Cos(π/4)
Cos(π/12)－Sin(π/12)=√2/2
設Cos(π/12)+Sin(π/12)=t>0
[Cos(π/12)+Sin(π/12)]^2+[Cos(π/12)－Sin(π/12)]^2=t^2+1/2
t^2+1/2=2
t^2=3/2=6/4
t=√6/2
Cos(π/12)+Sin(π/12)= √6/2
(e)
[Cos(π/12)+Sin(π/12)]+[Cos(π/12)－Sin(π/12)]= √6/2+√2/2
2 Cos(π/12)=√6/2+√2/2
Cos(π/12)=(√6+√2)/4
Sec(π/12)=4/(√6+√2)=4(√6－√2)/4=√6－√2