1
Question
Write equations whose roots are equal to numbers
cot2π2n+1+cot22π2n+1,cot23π2n+1,........,cot2nπ2n+1
cot2π2n+1+cot22π2n+1,cot23π2n+1,........,cot2nπ2n+1
Open in App
Solution
From Demoivre's theorem we know that
sin(2n+1)α=2n+1C1(cosα)2nsinα−2n+1C3(cosα)2n−2sin3α+.......+(−1)nsin2n+1α
or sin(2n+1)α=sin2n+1α{2n+1C1cot2nα−2n+1C3cot2n−2α+2n+1C5cot2n−4α−.....}
It follows that for
α=π2n+1,2π2n+1,3π2n+1,.......,nπ2n+1
Therefore equality holds
2n+1C1cot2nα−2n+1C3cot2n−2α+2n+1C5cot2n−4α−.......=0
It follows that the numbers
cot2π2n+1,cot22π2n+1,......,cot2nπ2n+1
are the roots of the equation
2n+1C1xn−2n+1C3xn−1+2n+1C5xn−2−.......=0 of nth degree.
sin(2n+1)α=2n+1C1(cosα)2nsinα−2n+1C3(cosα)2n−2sin3α+.......+(−1)nsin2n+1α
or sin(2n+1)α=sin2n+1α{2n+1C1cot2nα−2n+1C3cot2n−2α+2n+1C5cot2n−4α−.....}
It follows that for
α=π2n+1,2π2n+1,3π2n+1,.......,nπ2n+1
Therefore equality holds
2n+1C1cot2nα−2n+1C3cot2n−2α+2n+1C5cot2n−4α−.......=0
It follows that the numbers
cot2π2n+1,cot22π2n+1,......,cot2nπ2n+1
are the roots of the equation
2n+1C1xn−2n+1C3xn−1+2n+1C5xn−2−.......=0 of nth degree.
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
