Question

求证：sin1+sin2+sin3+......+sinn <2

Answer 1

证明：（1）2sin1(sin1+sin2+sin3+...+sinn)
=cos0-cos2+cos1-cos3+cos2-cos4+...
+cos(n-2)-cosn+cos(n-1)-cos(n+1)
=cos0+cos1-cosn-cos(n+1)
（2）=>S=sin1+sin2+sin3+...+sinn=[cos0+cos1-cosn-cos(n+1)]/2sin1
=>S=[1+cos1-(cosn+cos(n+1))]/[2sin(1/2)*cos(1/2)]
=>S=[2*cos(1/2)*cos(1/2)-2cos(n+1/2)*cos(1/2)]/[2sin(1/2)*cos(1/2)]
=>S=[cos(1/2)-cos(n+1/2)]/sin(1/2)

（3）因为-cos(n+1/2)最大为1

所以[cos(1/2)-cos(n+1/2)]/2sin(1/2)<[cos(1/2)+1]/sin(1/2)

（4）又因为[cos(1/2)+1]/sin(1/2)=1/tg(1/2)

注意到sinx<x<tgx (0<x<pi/2)
所以有1/tg(1/2)<1/(1/2)=2

（5）至此sin1+sin2+sin3+...+sinn=[cos(1/2)-cos(n+1/2)]/sin(1/2)<1/tg(1/2)<2