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 "

Question

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 "

Pratham Ashish · Answer

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&para;k x = e i2&para;k/n where k goes from 0 to n-1 x= 1, e i2&para;/n ,e i4&para;/n ,e i6&para;/n .............................e i2 (n-1)&para;/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&para;/n +e i4&para;/n +e i6&para;/n ............................+.e i2 (n-1)&para;/n =0 e i2&para;/n +e i4&para;/n +e i6&para;/n ............................+.e i2 (n-1)&para;/n = -1 hence proved (ii) sin n&Oslash; = b0 + b1 sin&Oslash; + b2 sin2&Oslash; + b3 sin3&Oslash;.................................+ bn sin n&Oslash; in this case , we can find the values of b0 ,b1, b2.......... by succesive differentiation of both the sides & then putting &Oslash; =0 , in that eq. for b0, put &Oslash; = 0 , we get b0 =0 for b1, on diff. both the sides & then putting &Oslash; = 0, we get b1 = n for furthur coeff. we have to do furthur diff.

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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 "

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