高考数学知识点-三角函数公式大全

高考数学知识点:三角函数公式大全sin =的对边 / 斜边cos =的邻边 / 斜边tan =的对边 / 的邻边cot =的邻边 / 的对边Sin2A=2SinA?CosACos2A=CosA2-SinA2=1-2SinA2=2CosA2-1tan2A=(2tanA)/(1-tanA2)(注：SinA2 是 sinA的平方 sin2(A) )sin3=4sinsin(/3+)sin(/3-)cos3=4coscos(/3+)cos(/3-)tan3a = tan a tan(/3+a) tan(/3-a)sin3a=sin(2a+a)=sin2acosa+cos2asinaAsin+Bcos=(A2+B2)(1/2)sin(+t)，其中sint=B/(A2+B2)(1/2)cost=A/(A2+B2)(1/2)tant=B/AAsin+Bcos=(A2+B2)(1/2)cos(-t)，tant=A/Bsin2()=(1-cos(2)/2=versin(2)/2cos2()=(1+cos(2)/2=covers(2)/2tan2()=(1-cos(2)/(1+cos(2)tan+cot=2/sin2tan-cot=-2cot21+cos2=2cos21-cos2=2sin21+sin=(sin/2+cos/2)2=2sina(1-sina)+(1-2sina)sina=3sina-4sinacos3a=cos(2a+a)=cos2acosa-sin2asina=(2cosa-1)cosa-2(1-sina)cosa=4cosa-3cosasin3a=3sina-4sina=4sina(3/4-sina)=4sina=4sina(sin60-sina)=4sina(sin60+sina)(sin60-sina)=4sina*2sincos*2sincos=4sinasin(60+a)sin(60-a)cos3a=4cosa-3cosa=4cosa(cosa-3/4)=4cosa=4cosa(cosa-cos30)=4cosa(cosa+cos30)(cosa-cos30)=4cosa*2coscos*-2sinsin=-4cosasin(a+30)sin(a-30)=-4cosasinsin=-4cosacos(60-a)=4cosacos(60-a)cos(60+a)上述两式相比可得tan3a=tanatan(60-a)tan(60+a)tan(A/2)=(1-cosA)/sinA=sinA/(1+cosA);cot(A/2)=sinA/(1-cosA)=(1+cosA)/sinA.sin2(a/2)=(1-cos(a)/2cos2(a/2)=(1+cos(a)/2tan(a/2)=(1-cos(a)/sin(a)=sin(a)/(1+cos(a)sin(+)=sincoscos+cossincos+coscossin-sinsinsincos(+)=coscoscos-cossinsin-sincossin-sinsincostan(+)=(tan+tan+tan-tantantan)/(1-tantan-tantan-tantan)cos(+)=coscos-sinsincos(-)=coscos+sinsinsin()=sincoscossintan(+)=(tan+tan)/(1-tantan)tan(-)=(tan-tan)/(1+tantan)sin+sin = 2 sin cossin-sin = 2 cos sincos+cos = 2 cos coscos-cos = -2 sin sintanA+tanB=sin(A+B)/cosAcosB=tan(A+B)(1-tanAtanB)tanA-tanB=sin(A-B)/cosAcosB=tan(A-B)(1+tanAtanB)sinsin = /2coscos = /2sincos = /2cossin = /2sin(-) = -sincos(-) = costan (a)=-tansin(/2-) = coscos(/2-) = sinsin(/2+) = coscos(/2+) = -sinsin() = sincos() = -cossin() = -sincos() = -costanA= sinA/cosAtan(/2+)=-cottan(/2-)=cottan()=-tantan()=tan诱导公式记背诀窍：奇变偶不变，符号看象限sin=2tan(/2)/cos=/1+tan(/2)tan=2tan(/2)/(1)(sin)2+(cos)2=1(2)1+(tan)2=(sec)2(3)1+(cot)2=(csc)2证明下面两式,只需将一式,左右同除(sin)2,第二个除(cos)2 即可(4)对于任意非直角三角形,总有tanA+tanB+tanC=tanAtanBtanC证:A+B=-Ctan(A+B)=tan(-C)(tanA+tanB)/(1-tanAtanB)=(tan-tanC)/(1+tantanC)整理可得tanA+tanB+tanC=tanAtanBtanC得证同样可以得证,当 x+y+z=nZ)时,该关系式也成立由 tanA+tanB+tanC=tanAtanBtanC可得出以下结论(5)cotAcotB+cotAcotC+cotBcotC=1(6)cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2)(7)(cosA)2+(cosB)2+(cosC)2=1-2cosAcosBcosC(8)(sinA)2+(sinB)2+(sinC)2=2+2cosAcosBcosC(9)sin+sin(+2/n)+sin(+2*2/n)+sin(