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  • 1 # s1985516s

    正弦,餘弦正切:首先推匯出兩角和公式:sin(x+y)=sinxcosy+cosxsiny 令x=θ/2,y=θ/2 sin(θ/2+θ/2)=sinθ/2cosθ/2+cosθ/2sinθ/2 得到:cosθ/2=sinθ/2sinθ/2 sin(x-y)=sinxcosy-cosxsiny 令x=θ,y=θ/2 sin(θ-θ/2)=sinθcosθ/2-cosθsinθ/2 sinθ/2=sinθ(sinθ/2sinθ/2)-cosθsinθ/2 sin²θ/2=sin²θ/2(1+cosθ) sin²θ/2=(1-cos²θ)/2(1+cosθ) sin²θ/2=(1+cosθ)/2 sinθ/2=±√(1+cosθ)/2 cos(a+b)=coacosb-sinasinb 令a=b=d cos2d=(cosd)^2-(sind)^2=(cosd)^2-[1-(cosd)^2]=2(cosd)^2-1 所以(cosd)^2=(cos2d+1)/2 以d/2代d,開方有cosd/2=±√[(1+cosd)/2] 而cos2d=(cosd)^2-(sind)^2=[1-(sind)^2]-(sind)^2=1-2(sind)^2 所以(sind)^2=(1-cos2d)/2 同樣的方法有sind/2=±√[(1-cosd)/2] tand/2=(sind/2)/(cosd/2)=±√[(1-cosd)/(1+cosd/2)] 還有一個是tand=sin2d/(1+cos2d)=(1-cos2d)/sin2d,推導如下: tand=sind/cosd=(2sindcosd)/(2cosdcosd)=sin2d/2(cosd)^2=sin2d/(1+cos2d) tand=sind/cosd=(2sindsind)/(2cosdsind)=2(sind)^2/sin2d=(1-cos2d)/sin2d [最後一步用了C(2d)的變形]

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