Question

高一数学公式，其中一个是这样sin(a+b)=sinacosb+cosasinb，求这类公式完整的

Answer 1

两角和公式 sin(A+B) = sinAcosB+cosAsinB sin(A-B) = sinAcosB-cosAsinB 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)
02
倍角公式 tan2A = 2tanA/(1-tan² A) Sin2A=2SinA•CosA Cos2A = Cos^2 A--Sin² A =2Cos² A—1 =1—2sin^2 A
03
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)
04
半角公式 sin(A/2) = √{(1--cosA)/2} cos(A/2) = √{(1+cosA)/2} tan(A/2) = √{(1--cosA)/(1+cosA)} cot(A/2) = √{(1+cosA)/(1-cosA)} ? tan(A/2) = (1--cosA)/sinA=sinA/(1+cosA)
05
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
06
积化和差 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)] cos(a)sin(b) = 1/2*[sin(a+b)-sin(a-b)]