半角公式的推导过程

正弦,余弦正切：首先推导出两角和公式：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,开方有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 同样的方法有sind/2=±√[(1-cosd)/2] tand/2=(sind/2)/(cosd/2)=±√[(1-cosd)/(1+cosd/2)] 还有一个是tand=sin2d/(1+cos2d)=(1-cos2d)/sin2d,推导如下： tand=sind/cosd=(2sindcosd)/(2cosdcosd)=sin2d/2(cosd)^2=sin2d/(1+cos2d) tand=sind/cosd=(2sindsind)/(2cosdsind)=2(sind)^2/sin2d=(1-cos2d)/sin2d [最后一步用了C(2d)的变形]