HMMT 二月 2010 · TEAM2 赛 · 第 7 题

英文解析

7. [ 30 ] Evaluate
   2010
   ∑
   cos ( k )2
   k =1
   sin(4021) − sin(1) 1+cos(2 x )
   Answer: 1005 + We use the identity cos ( k ) = . Then our expression 2
   4 sin(1) 2
   (cos(2)+ ... +cos(4020))
   evalutes to 1005 + .
   To evaluate cos(2) + · · · + cos(4020), let y = cos(2) + · · · + cos(4020) ⇒ y (sin(1)) = cos(2) sin(1) +2
   sin( x +1) − sin( x − 1)
   . . . + cos(4020) sin(1). Observe that for any x , cos( x ) sin(1) = . Then y (sin(1)) =
   sin(3) − sin(1) sin(5) − sin(3) sin 4021 − sin(4019)2 + + . . . + . This is clearly a telescoping sum; we get y (sin(1)) =
   2 2 2
   sin(4021) − sin(1) sin(4021) − sin(1)
   . Then we have the desired y = . Then our original expression evaluates
   2 2 sin(1)
   sin(4021) − sin(1)
   to 1005 + .
   4 sin(1)