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# Show that All Roots of ( X + 1 ) 6 + ( X − 1 ) 6 = 0 Are Given by -icot ( 2 K + 1 ) N 12 Where K=0,1,2,3,4,5. - Applied Mathematics 1

ConceptPowers and Roots of Trigonometric Functions

#### Question

Show that all roots of (x+1)^6+(x-1)^6=0 are given by -icot((2k+1)n)/12where k=0,1,2,3,4,5.

#### Solution

(x+1)^6+(x-1)^6=0

∴ (x+1)^6=(-x-1)^6

∴ (x+1)^6/(x-1)^6=-1

∴ ((x+1)/(x-1))^6=e^(ipi)         {∵ e^(ipi)=cospi+isinpi=-1+i(0)=-1}      (Principal value)

∴ ((x+1)/(x-1)^6=e^(i(pi+2kpi))         , k=0,1,2,3,4,5

∴ (x+1)/(x-1)=e(ipi(1+2k)/6)

Let 2θ= (pi(1+2k))/6

∴ From (1) & (2), (x+1)/(x-1)=e^(i2θ)

∴ By Componendo – Dividendo,( (x+1)+(x-1))/((x+1)-(x-1))=(e^12θ+1)/(e^12θ-1)

∴ 2x/2= (e^(iθ)[e^(iθ)+e^(-iθ)])/(e^(iθ)[e^(iθ)-e^(-iθ)])

∴ x=(2cosθ)/(2isinθ)      {∵sinθ=(e^(iθ)-e^(-iθ))/(2i) and cos θ= (e^(iθ)+e^(-iθ))/(2) }

∴ x=1/i cotθ

∴ x=-icot  ((2k+1)n)/12(From 2) where k = 0, 1, 2, 3, 4, 5

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Solution Show that All Roots of ( X + 1 ) 6 + ( X − 1 ) 6 = 0 Are Given by -icot ( 2 K + 1 ) N 12 Where K=0,1,2,3,4,5. Concept: Powers and Roots of Trigonometric Functions.
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