394 Algebra
tan(α+β+γ)=tanα+tanβ+tanγ−tanαtanβtanγ
1 −(tanαtanβ+tanβtanγ+tanαtanγ),
implies
tan(α+β+γ)=u+v+w−uvw
1 −(uv+vw+uv).
Using Viète’s relations,
u+v+w= 0 ,uv+vw+uw=− 10 ,uvw=− 11 ,we further transform this into tan(α+β+γ)= 1 +^1110 =1. Therefore,α+β+γ=π 4 +kπ,
wherekis an integer that remains to be determined.
From Viète’s relations we can see the product of the zeros of the polynomial is
negative, so the number of negative zeros is odd. And since the sum of the zeros is 0,
two of them are positive and one is negative. Therefore, one ofα, β, γlies in the interval
(−π 2 , 0 )and two of them lie in( 0 ,π 2 ). Hencekmust be equal to 0, and arctanu+
arctanv+arctanw=π 4.
Second solution: Because
Im ln( 1 +ix)=arctanx,we see that
arctanu+arctanv+arctanw=Im ln(iP (i))=Im ln( 11 + 11 i)
=arctan 1=π
4.
(K ̋ozépiskolai Matematikai Lapok(Mathematics Magazine for High Schools, Bu-
dapest), proposed by K. Bérczi).
162.Expanding the binomial(cosα+isinα)m, and using the de Moivre formula,
(cosα+isinα)m=cosmα+isinmα,we obtain
sinmα=(
m
1)
cosm−^1 αsinα−(
m
3)
cosm−^3 αsin^3 α+(
m
5)
cosm−^5 αsin^5 α+···.Form= 2 n+1, ifα= 2 nπ+ 1 , 2 n^2 π+ 1 ,..., 2 nπn+ 1 then sin( 2 n+ 1 )α=0, and sinαand cosα
are both different from zero. Dividing the above relation by sin^2 nα, we find that
(
2 n+ 1
1
)
cot^2 nα−(
2 n+ 1
3)
cot^2 n−^2 α+···+(− 1 )n(
2 n+ 1
2 n+ 1