求数学特殊角的三角函数公式、倍角公式。

Question

如题所述

Answer 1

sin(a+b)=sin(a)cos(b)+cos(α)sin(b)
根据这个推出来sin(2a)=2sin(a)cos(a)
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
根据这个推出来cos(2a)=(cosa)^2-(sina)^2=2(cosa)^2-1=1-2(sina)^2
sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
然后是乘除的，
根据sin(a+b)=sin(a)cos(b)+cos(α)sin(b)和sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
相加得到：(sin(a+b)+sin(a-b))/2=sin(a)cos(b)
其中如果使用(A+B)/2替换a,(A-B)/2替换b
就有sin(A)+sin(B)=2sin((A+B)/2)cos((A-B)/2)

Answer 2

sin(2a)=2sin(a)cos(a)
cos(2a)=cos^2(a)-sin^2(a)=2cos^2(a)-1=1-2sin^2(a)
sin(a+b)=sin(a)cos(b)+cos(α)sin(b)
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
和差化积、积化和差
sina+sinb=2sin[(a+b)/2]cos[(a-b)/2]
sina-sinb=2cos[(a+b)/2]2sin[(a-b)/2]
cos(a)
+cos(b)=2cos[(a+b)/2]cos[(a-b)/2]
cos(a)
-cos(b)=-2sin[(a+b)/2]sin[(a-b)/2]
sinasinb=-[cos(a+b)-cos(a-b)]/2
sinacosb=[sin(a+b)+cos(a-b)]/2
cosacosb=[cos(a+b)+cos(a-b)]/2
cosasinb=[sin(a+b)-cos(a-b)]/2
万能公式
sin2x=2tanx/(1+tanxtanx）
cos2x=(1+tanxtanx)/1-tanxtanx)
tan2x=2tanx/(1+tanxtanx)
tan(a/2）=[1-cosa]/sina=sina/(1+cosa)
基本差不多了，常用的。

Answer 3

二倍角公式
sin(2a)=2sin(a)cos(a)
cos(2a)=cos^2(a)-sin^2(a)=2cos^2(a)-1=1-2sin^2(a)
两角和与差的三角函数
sin(a+b)=sin(a)cos(b)+cos(α)sin(b)
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
cos(a-b)=cos(a)cos(b)+sin(a)sin(b)

Answer 4

二倍角公式
sin(2a)=2sin(a)cos(a)
cos(2a)=cos^2(a)-sin^2(a)=2cos^2(a)-1=1-2sin^2(a)
两角和与差的三角函数
sin(a+b)=sin(a)cos(b)+cos(α)sin(b)
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
cos(a-b)=cos(a)cos(b)+sin(a)sin(b)

Answer 5

我来补充一个外能公式吧
记t=tanx
则tan2x=2t/(1-t的平方)
sin2x=2t/(1+t的平方)
cos2x=(1+t的平方)/(1-t的平方)