三角函数万能公式的推导

设tan(A/2)=t
sinA=2t/(1+t^2)
tanA=2t/(1-t^2)
cosA=(1-t^2)/(1+t^2)

推导第一个： (其它类似)
sinA=2sin(A/2)cos(A/2)
=[2sin(A/2)cos(A/2)]/[sin^2(A/2)+cos^2(A/2)]
分子分母同时除以cos^2(A/2)
=[2sin(A/2)cos(A/2)/cos^2(A/2)]/[(sin^2(A/2)+cos^2(A/2))/cos^2(A/2)]
化简：
=[2sin(A/2)/cos(A/2)]/[sin^2(A/2)/cos^2(A/2)+1]
即：
=(2tan(A/2))/(tan^(A/2)+1)

sin2A=2sinAcosA=2sinAcosA/(cos^2A+sin^2A)......*,(因为cos^2A+sin^2A=1),再把*分式上下同除cos^2A,可得
余弦的也是化为二倍角,除以cos^2A+sin^2A
http://dl.zhishi.sina.com.cn/upload/62/84/00/1253628400.11366413.jpg

sinα=2sin(α/2)cos(α/2)
=[2sin(α/2)cos(α/2)]/[sin(α/2)^2+cos(α/2)^2]
=[2tan(α/2)]/[1+(tanα/2)^2]

cosα=[cos(α/2)^2-sin(α/2)^2]
=[cos(α/2)^2-sin(α/2)^2]/[sin(a/2)^2+cos(a/2)^2]
=[1-tan(α/2)^2]/[1+(tanα/2)^2]

tanα=tan[2*(α/2)]
=2tan(α/2)/[1-tan(α/2)^2]
=[2tan(a/2)]/[1-(tanα/2)^2]