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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 "


 


 


 

10 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.

 

 

10 years ago
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