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[SPAM] Re: [obm-l] trigonometria
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--0-1960570269-1212469275=:44292
Content-Type: text/plain; charset=utf-8
Content-Transfer-Encoding: quoted-printable
=20
Seja z =3D Pi {k=3D1-> 89){sen kx} onde x=3D1=
=C2=BA e Pi (k=3D1->m) =C3=A9 o produt=C3=B3rio para k variando &=
nbsp; de 1 a m (natuiais, naturalmente....he he he..).
=20
z^2 =3D Pi (k=3D1->89){(sen kx)^2 =3D (sen 45=C2=BA)^=
2 * Pi(k=3D1->44} {(sen kx)^2 *[1 - (sen kx)^2]},=20
j=C3=A1 que sen (90 -kx) =3D cos kx.
..
Denominando y(k) =3D (sen kx)^2 , temos, z^2 =3D(1/2) Pi (k=3D=
1->44) {y(k) -(y(k))^2} com =20
y(k)<1/2 portanto y(k) -(y(k))^2 < 1/4, para k<45.
Assim z^2 < (1/2)*1/2^(2*44) ou z < 1/2^(44,5) . &=
nbsp; como z =3D 1/2^n , n> 44,5 e n=C3=A3o menor que 45
--- Em qui, 29/5/08, Pedro J=C3=BAnior <pedromatematico06@xxxxxxxxx> =
escreveu:
De: Pedro J=C3=BAnior <pedromatematico06@xxxxxxxxx>
Assunto: [obm-l] trigonometria
Para: "obm-l" <obm-l@xxxxxxxxxxxxxx>
Data: Quinta-feira, 29 de Maio de 2008, 23:22
Boa noite a todos...
Me deparei com esse probleminha e ainda n=C3=A3o consegui v=C3=AA a sa=C3=
=ADda!
Sabendo-se que sen1=C2=B0 .sen2=C2=B0. sen3=C2=B0 . ... . sen85=C2=B0 .sen8=
7=C2=B0 .sen89=C2=B0 =3D 1/2^n, mostre que n<45.
Acho que algu=C3=A9m mandou e minha esposa limpou miha caixa de e-mail's e =
a solu=C3=A7=C3=A3o foi junto, parece piada, mas foi o que aconteceu!
Se algu=C3=A9m tiver enviado e puder enviar novamente agrade=C3=A7o desde j=
=C3=A1!
Abra=C3=A7o a todos.
=0A=0A=0A Abra sua conta no Yahoo! Mail, o =C3=BAnico sem limite de es=
pa=C3=A7o para armazenamento!=0Ahttp://br.mail.yahoo.com/
--0-1960570269-1212469275=:44292
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: quoted-printable
<table cellspacing=3D'0' cellpadding=3D'0' border=3D'0' background=3D'none'=
style=3D'font-family:arial;font-size:10pt;color:rgb(51, 51, 51);background=
-color:rgb(255, 255, 255);width:100%;'><tr><td valign=3D'top' style=3D'font=
: inherit;'> <br> Seja z =3D Pi {k=3D1-> 89){sen kx}&n=
bsp; onde x=3D1=C2=BA e Pi (k=3D1->m) =C3=A9 o produt=C3=B3rio par=
a k variando de 1 a m (natuiais, naturalmente....=
he he he..).<br> <br> z^2 =3D Pi (k=3D1->89){(se=
n kx)^2 =3D (sen 45=C2=BA)^2 * Pi(k=3D1->44} {(sen kx)^2 *[1 - (se=
n kx)^2]}, <br> j=C3=A1 que sen (90 -kx) =3D cos kx.<br>.=
.<br> Denominando y(k) =3D (sen kx)^2 , temos, z^2 =3D(1/2) Pi =
(k=3D1->44) {y(k) -(y(k))^2} com <br> y(k)<1/2&nbs=
p; portanto y(k) -(y(k))^2 < 1/4, para k<45.<br><br> Assim z^2 =
< (1/2)*1/2^(2*44) ou z < 1/2^(44,5) . como z =3D =
1/2^n , n> 44,5 e n=C3=A3o menor que 45<br><hr
style=3D"width: 100%; height: 2px;"><br><br><br>--- Em <b>qui, 29/5/08, Pe=
dro J=C3=BAnior <i><pedromatematico06@xxxxxxxxx></i></b> escreveu:<br=
><blockquote style=3D"border-left: 2px solid rgb(16, 16, 255); margin-left:=
5px; padding-left: 5px;">De: Pedro J=C3=BAnior <pedromatematico06@gmail=
.com><br>Assunto: [obm-l] trigonometria<br>Para: "obm-l" <obm-l@xxxxx=
uc-rio.br><br>Data: Quinta-feira, 29 de Maio de 2008, 23:22<br><br><div =
id=3D"yiv939910206">Boa noite a todos...<br>Me deparei com esse probleminha=
e ainda n=C3=A3o consegui v=C3=AA a sa=C3=ADda!<br><br>Sabendo-se que sen1=
=C2=B0 .sen2=C2=B0. sen3=C2=B0 . ... . sen85=C2=B0 .sen87=C2=B0 .sen89=C2=
=B0 =3D 1/2^n, mostre que n<45.<br><br>Acho que algu=C3=A9m mandou e min=
ha esposa limpou miha caixa de e-mail's e a solu=C3=A7=C3=A3o foi junto, pa=
rece piada, mas foi o que aconteceu!<br>
Se algu=C3=A9m tiver enviado e puder enviar novamente agrade=C3=A7o desde j=
=C3=A1!<br>Abra=C3=A7o a todos.<br><br>
</div></blockquote></td></tr></table><br>=0A=0A=0A <hr size=3D1>Abra s=
ua conta no <a href=3D"http://br.rd.yahoo.com/mail/taglines/mail/*http://br=
.mail.yahoo.com/">Yahoo! Mail</a>, o =C3=BAnico sem limite de espa=C3=A7o p=
ara armazenamento! =0A
--0-1960570269-1212469275=:44292--
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