数学恒等式

11111xcoth(x)1

......12x222x232x242x252x22x211111xcot(x)1

......12x222x232x242x252x22x2111112sinh(2x)sin(2x)144444444......3441x2x3x4x5x4xcosh(2x)cos(2x)2x42，证明：



nn11cossin(2rsin)sinsinh(2rcos)142r2(cos2)n2r44r3sin2sinh2(rcos)sin2(rsin)2r4

153,设

234证明：t5s12s4s23s3s4

xxs,txx51110x2x11x3x1511x4x2011x5x25111......1......13s4s2ss4,证明：

e25e21e41e61e811......55151 225,n是一个正整数，证明：

031323332n3nnn(C2)(C)(C)(C)...(C)(1)CCn2n2n2n2n3n2n

6，证明：

e0tx2dx22tt2tet2

123t2t45t......7,三角形ABC三边长分别是x+y,y+z,z+x,内切圆I分别与三边BC,CA,AB相切于D,E,F，AD,BE,CF三线共点，它们相交于J，证明IJ的长度等于： (x1y1z1)1(xyz)1(xyz1xy1zx1yzxyz) 8,证明：

0dx 222423610(1x)(1tx)(1tx)......2(1tttt......)