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# Please show proper notation, justification and step by step work.

Please show proper notation, justification and step by step work. n See attachment for problem
Given that the zeros for (sinx)/x are the values  x=0, x= +-pie, x=+-2pie, x=+-3pie,. (x=0 must be excluded, why?)
This implies that F(x) can be factored as follows
F(x) = (1-(x/pi))  (1-(x/-pi))  (1-(x/2pi))  (1-(x/3pi))  (1-(x/-3pi)).
= [(1-(x/pi))(1+(x/pi))]  [(1-(x/2pi))(1+(x/2pi))]  [(1-(x/3pi))(1+(x/3pi)).
= (1-((x^2)/(pi^2)))  (1-((x^2)/(4pi^2)))  (1-((x^2)/(9pi^2)))  (1-((x^2)/(16pi^2)))  (1-  ((x^2)/(25pi^2)))
Multiply out this infinite product (or at least the first 6 factors) and collect like terms in x's to form a polynomial.
i.e   F(x)=  1- (???????)x^2 + (????????)x^4  -(???????)x^6  +..

My notation is very bad but it is the best I could do with my version of
windows. Please show step by step work and a nice tidy answer.
Given that the zeros for (sinx)/x are the values x=0, x= +-pie,
x=+-2pie, x=+-3pie,â¦. (x=0 must be excluded, why?)
This implies that F(x) can be factored as follows
F(x) = (1-(x/pi)) (1-(x/-pi)) (1-(x/2pi)) (1-(x/3pi))
(1-(x/-3pi))â¦.
= [(1-(x/pi))(1+(x/pi))] [(1-(x/2pi))(1+(x/2pi))]
[(1-(x/3pi))(1+(x/3pi))â¦.
= (1-((x^2)/(pi^2))) (1-((x^2)/(4pi^2))) (1-((x^2)/(9pi^2)))
(1-((x^2)/(16pi^2))) (1- ((x^2)/(25pi^2)))â¦
Multiply out this infinite product (or at least the first 6 factors) and
collect like terms in xâs to form a polynomial.
i.e F(x)= 1- (???????)x^2 + (????????)x^4 -(???????)x^6 +â¦â¦..

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