三角函数的转换关系

题目

三角函数的转换关系

答案

两角和公式
sin(A+B) = sinAcosB+cosAsinB
sin(A-B) = sinAcosB-sinBcosA 
cos(A+B) = cosAcosB-sinAsinB
cos(A-B) = cosAcosB+sinAsinB
tan(A+B) = (tanA+tanB)/(1-tanAtanB)
tan(A-B) = (tanA-tanB)/(1+tanAtanB)
cot(A+B) = (cotAcotB-1)/(cotB+cotA) 
cot(A-B) = (cotAcotB+1)/(cotB-cotA)
倍角公式
tan2A = 2tanA/[1-(tanA)²]
cos2a = (cosa)²-(sina)²=2(cosa)² -1=1-2(sina)²
sin2A = 2sinA·cosA
三倍角公式
sin3a = 3sina-4(sina)³
cos3a = 4(cosa)³-3cosa
tan3a = tan a · tan(π/3+a)· tan(π/3-a)
半角公式
sin(A/2) = √((1-cosA)/2) sin(A/2)=-√((1-cosA)/2)
cos(A/2) = √((1+cosA)/2) cos(A/2)=-√((1+cosA)/2)
tan(A/2) = √((1-cosA)/((1+cosA)) tan(A/2)=-√((1-cosA)/((1+cosA))
cot(A/2) = √((1+cosA)/((1-cosA)) cot(A/2)=-√((1+cosA)/((1-cosA)) 
tan(A/2) = (1-cosA)/sinA=sinA/(1+cosA)
和差化积
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]
tanA+tanB=sin(A+B)/cosAcosB
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)]
诱导公式
sin(-a) = -sin(a)
cos(-a) = cos(a)
sin(π/2-a) = cos(a)
cos(π/2-a) = sin(a)
sin(π/2+a) = cos(a)
cos(π/2+a) = -sin(a)
sin(π-a) = sin(a)
cos(π-a) = -cos(a)
sin(π+a) = -sin(a)
cos(π+a) = -cos(a)
tgA=tanA = sinA/cosA
万能公式
sin(a) = [2tan(a/2)]/[1+tan²(a/2)]
cos(a) = [1-tan²(a/2)]/[1+tan²(a/2)]
tan(a) = [2tan(a/2)]/[1-tan²(a/2)]
其它公式
a·sin(a)+b·cos(a) = sqrt(a²+b²)sin(a+c) [其中,tan(c)=b/a]
a·sin(a)-b·cos(a) = sqrt(a²+b²)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