三角函数的转换关系

Question

如题所述

Answer 1

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