how to solve the equation (1+x)^(2n)+(1-x)^(2n)=0
Answers
Answered by
0
(1+x)^2n + (1-x)^2n=0
or, {(1+x)^n}^2 + {(1-x)^n}^2 =0
Is it a theorem that if the value of addition of some number's square is 0, then the value of the number's square will differently 0.
.'. {(1+x)^n}^2 = 0 and {(1-x)^n}^2 = 0
or, (1+x)^n =0 and (1-x)^n =0
or, (1+x)=0 and (1-x)=0
or, x=-1 and x=1
or, {(1+x)^n}^2 + {(1-x)^n}^2 =0
Is it a theorem that if the value of addition of some number's square is 0, then the value of the number's square will differently 0.
.'. {(1+x)^n}^2 = 0 and {(1-x)^n}^2 = 0
or, (1+x)^n =0 and (1-x)^n =0
or, (1+x)=0 and (1-x)=0
or, x=-1 and x=1
RupkumarDolai:
(1+x)^2n/(1-x)^2n= -1
ok
ok
now we can write -1 into this form.. cos(2k-1)π+i sin(2k+1)π
ok
putting k= 1,2,... ,n you always get -1
now here's a theorem its call De moivre's theorem
yes,yes
now, (1+x)^2n/(1-x)^2n=cos(2k-1)π+i sin(2k-1)π this implied (1+x)/(1-x)= cos(2k-1)π/2n+i sin(2k-1)π/2n
by de moivre's theorem
Similar questions