arctan2-arctan0

设arctan1=t,arctan2=s,arctan3=k,则tant=1,tans=2,tank=3,tan(s+k)=(2+3)/(1-2*3)=-1,于是tan(t+s+k)=(tant+ta

证明:设arctan1+arctan2+arctan3=x那么tanx=tan(arctan1+arctan2+arctan3)=(tan(arctan1+arctan2)+tan(arctan3))

设arctan1+arctan2+arctan3=x那么tanx=tan(arctan1+arctan2+arctan3)=(tan(arctan1+arctan2)+tan(arctan3))/(1

设arctan1=t,arctan2=s,arctan3=k,则tant=1,tans=2,tank=3,tan(s+k)=(2+3)/(1-2*3)=-1,于是,tan(t+s+k)=(tant+t

设tanA=1.tanB=2,tanC=3,D=A+BtanD=tan（A+B）=(1+2)/(1-1*2)=-3tan(A+B+C)=tan(D+C)=(-3+3)/(1+9)=0=>A+B+C=1

设tanA=1.tanB=2,tanC=3,D=A+BtanD=tan（A+B）=(1+2)/(1-1*2)=-3tan(A+B+C)=tan(D+C)=(-3+3)/(1+9)=0=>A+B+C=1

因为arctan1=π/4只要证明arctan2+arctan3=3π/4即可,因为tan（arctan2+arctan3）=（2+3）/（1-2*3）=-1又π/4

设arctan1=A,arctan2=B,arctan3=C则tanA=1,tanB=2,tanC=3tan(A+B)=(tanA+tanB)/（1-tanAtanB)=(1+2)/(1-2)=-3=

两边同时取正切tan(arctan2/x+arctan3/x)=tan(45°)(tan(arctan2/x)+tan(arctan3/x))/(1-(tan(arctan2/x)*tan(arcta

等价于0.8x+tanx=0x的范围是0.5pi~1.5pi就一个解,二分法~再问：请问能不能再说的详细下还是不懂呵呵再答：额~~实在懒得写。x+（arctan0.8）x=3.14arctan0.8x

∫(1-cosx)(sinx)^2dx=∫(sinx)^2dx-∫(sinx)^2dsinx=∫(1/2-cos2x/2)dx-(sinx)^3/3=x/2-sin2x/4-(sinx)^3/3又si

11.309932474度.约等于11.31度即11度18分36秒.

63.43

（2sinx-cosx）^2=1;4(sinx)^2+(cosx)^2-4sinxconx=1;3(sinx)^2-4sinxcosx=0,结合2sinx+1=cosx;可得答案

A=2T=8w=2π/8=π/4x=2时,f(x)max=2得：f(x)=2sinπ/4xtan倾斜角=tan（y/x）=2,得:倾斜角为arctan2不等式为(2^x+1)*(2^x)-2*1>=0

设a=arctan3,则tana=3b=arctan2,则tanb=2tan(arctan3-arctan2)=tan(a-b)=(tana-tanb)/(1+tanatanb)=1/(1+6)=1/

tanπ/4=1tan0=0所以原式=π/4-0=π/4

arctan1等于π/4,即45°,arctan0等于0,即0°.

§(1-cosx)sin2xdx=1/2(x-sinxcosx)-sin^3x/3然后用牛顿～莱布尼茨公式即可