高二数学公式整理

高二数学公式整理高二数学公式：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=/）sin3=4sin·sin(/3+)sin(/3-)cos3=4cos·cos(/3+)cos(/3-)tan3a=tana·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(+)=sin·cos·cos+cos·sin·cos+cos·cos·sin-sin·sin·sincos(+)=cos·cos·cos-cos·sin·sin-sin·cos·sin-sin·sin·costan(+)=(tan+tan+tan-tan·tan·tan)/(1-tan·tan-tan·tan-tan·tan)cos(+)=cos·cos-sin·sincos(-)=cos·cos+sin·sinsin(±)=sin·cos±cos·sintan(+)=(tan+tan)/(1-tan·tan)tan(-)=(tan-tan)/(1+tan·tan)sin+sin=2sincossin-sin=2cossincos+cos=2coscoscos-cos=-2sinsintanA+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/cosAtancottancottantantantan万能公式sin=2tan(/2）/1+tan(/2)cos=1-tan(/2)/1+tan(/2)tan=2tan(/2)/1-tan(/2)其它公式(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=n(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)2+2+2=2+2cosAcosBcosC(9)sin+sin(+2/n)+sin(+2*2/n)+sin(+2*3/n)+sin=0cos+cos(+2/n)+cos(+2*2/n)+cos(+2*3/n)+cos=0以及sin2()+sin2(-2/3)+sin2(+2/3)=3/2tanAtanBtan(A+B)+tanA+tanB-tan(A+B)=0