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shakya sur Grade: 11
        

1)Why is it that


                 sigma [ k=1 to (n-1) ] cos [2k(pi)/n] = -1 ????                     


 


2) Could you suggest any approach for this one ? 


                  let n be an odd integer. If


                              sin [n(theta)] = sigma [ r=0 to n ] b (subscript r ) [sin (theta)]^r


                  then find the value of "b subscript 0 " and "b subsript 1 "


 


 


 

8 years ago

Answers : (1)

Pratham Ashish
17 Points
										

 hi,


consider the eq.


              x n -1 = 0


this eq. will have n roots,  which will be the n th roots of unity,


   x = 1


x n  =   e i2¶k


x =  e i2¶k/n


where k goes from 0  to n-1


 x=   1,  e i2¶/n  ,e i4¶/n ,e i6¶/n .............................e i2 (n-1)¶/n


from the analysis of eq . we can tell that sum of all the roots will be 0 ,becoz the coeff. of  x^(n-10 is zero here


so,


 1+e i2¶/n  +e i4¶/n +e i6¶/n ............................+.e i2 (n-1)¶/n    =0


e i2¶/n  +e i4¶/n +e i6¶/n ............................+.e i2 (n-1)¶/n    = -1  hence proved


 


 


(ii)


 sin nØ =  b0  + b1 sinØ  + b2  sin2Ø + b3  sin3Ø.................................+ bn sin nØ


in this case , we can find the values of b0 ,b1, b2..........  by succesive differentiation of both the


sides & then putting  Ø =0 , in that eq.


  for b0,


  put Ø  = 0 , we get


b0 =0


for b1,


on diff. both the sides  & then putting Ø = 0, we get


 b1 = n


  for furthur coeff. we have to do furthur diff.


 


 

8 years ago
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