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# 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 "

12 years ago

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.