# Sin x per x

Buktikan dengan cermat bahwa

untuk setiap x ≠ 0.

Catatan:

*

Bandung, 12-11-2016

1. karena sinx = 2cos(x/2)sin(x/2), jadi kita punya
sinx = 2^2 cos(x/2)cos(x/4)sin(x/4)
sinx = 2^3 cos(x/2)cos(x/4)cos(x/8)sin(x/8)
sinx = 2^4 cos(x/2)cos(x/4)cos(x/8)cos(x/16)sin(x/16)
…..
sinx = 2^N cos(x/2)cos(x/4)cos(x/8)cos(x/16)…cos(x/2^n)sin(x/2^n)
(sinx)/x = (2^N)/x sin(x/2^N) [pi n=1 to N (cos(x/2^n))]
lim n-> infinity [(sinx)/x [sin(x/2^N)/(x/2^N)]]^(-1)] = lim N-> infinity [pi n=1 to N (cos(x/2^n))]
(sinx)/x = lim N-> infinity [pi n=1 to N (cos(x/2^k))]
(sinx)/x = pi n=1 to infinity [cos(x/2^n], untuk setiap x bukan nol. QED

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2. karena sinx = 2cos(x/2)sin(x/2), jadi kita punya
sinx = 2^2 cos(x/2)cos(x/4)sin(x/4)
sinx = 2^3 cos(x/2)cos(x/4)cos(x/8)sin(x/8)
sinx = 2^4 cos(x/2)cos(x/4)cos(x/8)cos(x/16)sin(x/16)
…..
sinx = 2^N cos(x/2)cos(x/4)cos(x/8)cos(x/16)…cos(x/2^n)sin(x/2^n)
(sinx)/x = (2^N)/x sin(x/2^N) [pi n=1 to N (cos(x/2^n))]
lim n-> infinity [(sinx)/x [sin(x/2^N)/(x/2^N)]]^(-1)] = lim N-> infinity [pi n=1 to N (cos(x/2^n))]
(sinx)/x = lim N-> infinity [pi n=1 to N (cos(x/2^k))]
(sinx)/x = pi n=1 to infinity [cos(x/2^n], untuk setiap x bukan nol. QED

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