急求高一所有三角恒等式变换.

Question

急求高一所有三角恒等式变换.

Answer 1

诱导公式
sin(-a)=-sin(a)
cos(-a)=cos(a)
sin(pai/2-a)=cos(a)
cos(pai/2-a)=sin(a)
sin(pai/2+a)=cos(a)
cos(pai/2+a)=-sin(a)
sin(pai-a)=sin(a)
cos(pai-a)=-cos(a)
sin(pai+a)=-sin(a)
cos(pai+a)=-cos(a)
tgA=tanA=sinA/cosA
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)
tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b))
tan(a-b)=(tan(a)-tan(b))/(1+tan(a)tan(b))
和差化积公式
sin(a)+sin(b)=2sin((a+b)/2)cos((a-b)/2)
sin(a)−sin(b)=2cos((a+b)/2)sin((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)
积化和差公式
sin(a)sin(b)=-1/2*[cos(a+b)-cos(a-b)]
cos(a)cos(b)=1/2*[cos(a+b)+cos(a-b)]
sin(a)cos(b)=1/2*[sin(a+b)+sin(a-b)]
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)
半角公式
sin^2(a/2)=(1-cos(a))/2
cos^2(a/2)=(1+cos(a))/2
tan(a/2)=(1-cos(a))/sin(a)=sin(a)/(1+cos(a))
万能公式
sin(a)= (2tan(a/2))/(1+tan^2(a/2))
cos(a)= (1-tan^2(a/2))/(1+tan^2(a/2))
tan(a)= (2tan(a/2))/(1-tan^2(a/2))
其它公式
a*sin(a)+b*cos(a)=sqrt(a^2+b^2)sin(a+c) [其中,tan(c)=b/a]
a*sin(a)-b*cos(a)=sqrt(a^2+b^2)cos(a-c) [其中,tan(c)=a/b]
1+sin(a)=(sin(a/2)+cos(a/2))^2
1-sin(a)=(sin(a/2)-cos(a/2))^2
其他非重点三角函数
csc(a)=1/sin(a)
sec(a)=1/cos(a)
双曲函数
sinh(a)=(e^a-e^(-a))/2
cosh(a)=(e^a+e^(-a))/2
tgh(a)=sinh(a)/cosh(a)