融合 的 "M.I." 有一定難過

use MI

let P(n) be the statement
(cos2X+cos4X+cos6X.. ....+cos2nX) = sin(nX) cos(n+1)X/sinX

when n=1

LHS=cos2X
RHS=cos2X*sinX/sinX=cos2X

the statement is true

when n=k let P(k) is true, that is
(cos2X+cos4X+cos6X.. ....+cos2kX) = sin(kX) cos(k+1)X/sinX

when n=k+1
(cos2X+cos4X+cos6X......+cos2kX+cos2(k+1)X)
=[sin(kX) cos(k+1)X]/sinX+cos(2(k+1)X)
=1/sinX{[sin(kX) cos(k+1)X]+cos(2(k+1)X)sinX}
=(1/2sinX)(sin(-x)+sin(2k+1)X+sin(2k+3)X+sin(-(2k+1)X))
=(1/2sinX)(sin(2k+3)X+sin(-X))
=sin((k+1)X)cos(k+2)X/sinX
when n=k+1 P(k+1) is true

by MI ,for all values of n, P(n) is true
2
cos2X+cos4X+...+cos1 0X
=(cos(5+1)X sin 5X)/sinX (n=5)
=(cos6X sin 5X)/sinX
cos2X+cos4X+...+cos1 0X=0
(cos6X sin 5X)/sinX=0
(cos6X sin 5X)=0
cos6X=0 或 sin 5X=0
6X=90,270,450 或 5X=0,180,360
X=15,45,90 或 X=0,36,72
X=0,15,36,45,72,90