數學 M2 三角學-－vbd391t的部落格

標題:

數學 M2 三角學?

發問:

求37所有分題，請詳細列式

最佳解答:

(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