三角恒等變換公式大全

1、三角函數(shù)cos(a+ )=coSa' cos -Sina Sincos(a-)=cos a cos +Sina SinSin(a+ )=Sin a' cos +cosa SinSin(a-)=Sin a cos -cosI a Sintan(a+ )=(ta n a+ta n)/(1-tana tan)tan(a-)=(ta n a-ta n)/(1+ta na tan)二倍角Sin (2a) =2sin a cos a =2tan (a)/1-tan2(a)cos (2 a) =cos2 (a) -Si n2 (a) =2cos2 (a)-1=1-2si n2(a)=1-ta

2、 n2(a)1+tan2(a)tan (2a) =2tana /1 -tan2(a)三倍角sin3 =3sin -4sin3 (a)cos3 a =4cos3 (a) - 3COS atan3 a = (3tan a -tan3 (a)÷( 1-3tan2 (a)sin3 a =4sin a× Sin ( 60- a) Sin (60+a)cos3 a =4cos a× cos ( 60- a) cos ( 60+a)tan3 a =tan a× tan ( 60- a) tan (60+a)半角公式Sin 2(a /2 )=(1-cos a) /2co

3、s2(a /2 )=(1+cos a) /2tan 2(a /2 )=(1-cos a) / ( 1+cos a)tan (a /2 ) =Sina / ( 1+cos a) = ( 1- cos a) /si n a半角變形sin2 (a /2 ) = (1-cosa) /2Sin(a2 ) = (I-COSa) /2 a/2 在一、二象限=- (1-cos) /2 a/2 在三、四象限COS2 (a /2 ) = (1+cos ) /2cos(a2 ) = (1+cos ) /2 a/2 在一、四象限=- (1+cos ) /2 a/2 在二、三象限tan 2 ( /2 ) = (1- c

4、os a) / (1+cos a)tan (a /2 ) =S in a/ (1+cos a) =( 1- cos a) /si n a = (1- cos a) / (1+cos a)a/2在一、三象限=- (1- cos a) / (1+cos a) a/2 在二、四象限恒等變形tan(a+ /4 ) =(tana+1 ) / (1-tana)tan (a- /4 ) =(tana-1 ) / (1+tana)asinx+bcosx= ( a2+b2) a/ ( a2+b2) sinx+b (a2+b2) cosx= ( a2+b2) sin(x+y)(輔助角公式)tan y=ba萬能代換

5、半角的正弦、余弦和正切公式(降幕擴角公式)Sin =2tan ( /2 ) /1+tan2 ( /2 )cos a =1 -tan (a /2 ) /1+ta n2 (a /2 )tan a =2ta n (a /2 ) /1-ta n2 (a /2 )積和化差Sin a cos =:(1/2) sin(a + ) +si n(a-)cos a Sin =:(1/2) sin(a + ) -Si n(a-)cos a cos =:(1/2) cos(a + ) +cos(a- )Sin a Sin =:-(1/2 ) cos (a + ) -cos (a - ）（注：留意最前面是負號）和差化積

6、Sin a +sin =2sin(a + )/2cos(a -)/2Sin a -Sin =2cos(a + )/2s in(a -)/2cos a +cos =2cos(a + )/2cos(a -)/2cos a - cos =-2sin(a + )/2si n(a-)/2內(nèi)角公式SinA+sinB+sinC=4cos (A/2) COS ( B/2) COS (C/2) cosA+cosB+cosC=1+4sin (A/2) Sin (B/2) Sin (C/2) tan A+ta nB+ta nC=ta nAta nBta nCcot (A/2) +cot (B/2) +cot (C/

7、2) =cot (A/2) cot ( B/2) cot (C/2) tan( A/2)ta n(B2)+ta n(B2)ta n(C2)+ta n(C2)ta n(A2)=1COtACOtB+cotBcotC+cotCcotA=1證明方法首先，在三角形ABC中,角A,B,C所對邊分別為a,b,c若A,B均為銳角，貝恠三 角形ABC中，過C作AB邊垂線交AB于D由CD=asinB=bsinA(做另兩邊的垂線， 同理)可證明正弦定理：a/sinA=bsinB=csinC 于是有：AD+BD=c AD=bcosA,BD=acosB AD+BD=代入正弦定理，可得 SinC=sin(180-C)=Sin(A+B)=sinAcosB+sinBcosA 即在A,B均為銳角的情況下，可證明正 弦和的公式。利用正弦和余弦的