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发表于 29-10-2011 04:21 PM
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further maths T past year 来的哦~
我就做第六题,第八题很难type....
我let x = tita
assume the statement is true for all value is set of positive integers
summation (1,n) = (sin(2n+1)x)/(2sinx) - 1/2
summation(1,n+1) = (sin(2n+1)x)/(2sinx) - 1/2 + cos2nx
= (sin(2n+1)x)/(2sinx) - 1/2 + (2sinxcos2nx)/2sinx
= ((sin(2n+1)x)+(2sinxcos2nx))/(2sinx) - 1/2 ,
note that 2sinxcos2nx = sin[2nx+3x]-sin[2nx+x] compound angle formula...
hence,
((sin(2n+1)x)+(sin[2nx+3x]-sin[2nx+x])/(2sinx) - 1/2
= sin[2nx+3x]/(2sinx) - 1/2
= (sin[2(n+1)+1]x)/(2sinx) - 1/2
second part 会吧? |
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