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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 "

```
8 years ago

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 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.

```
8 years ago
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