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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 "
hi, consider the eq. x n -1 = 0 this eq. will have n roots, which will be the n th roots of unity, x n = 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.
hi,
consider the eq.
x n -1 = 0
this eq. will have n roots, which will be the n th roots of unity,
x n = 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.
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