koreankid36 I'm not sure if I'm correct because I forgot most of my trig... but.. here I go o-o: Lnlsec(x)l = Lnl1/cosxl = Lnl1l- Lnlcosxl (and because ln1=0) = -Lnlcosxl yup ... I think that's how to do it.. Reply January 6, 2011
jonatan sec(x) = 1/cos(x) Lnlsec(x)l=-Lnlcos(x)l Gives us:I: e^(Lnlsec(x)l)II: e^(-Lnlcos(x)l) Next step:I: e^(Lnlsec(x)l)=|sec(x)|=|1/cos(x)|=1/|cos(x)|II: e^(-Lnlcos(x)l) = 1/e^(Lnlcos(x)l) = 1/|cos(x)| I = II Maybe something like that? Reply January 6, 2011
Xgo321 :o Omg is that all? seems so short.. D: Ahh wtheck can't work with those numbers. Reply January 6, 2011
darksuitguy [quote=KidZombie]Those arent numbers.[/quote] You need to learn math.Ln means natural log (log base e), while lxl means absolute value of x. Reply January 6, 2011 - edited
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I'm not sure if I'm correct because I forgot most of my trig... but.. here I go o-o:
Lnlsec(x)l = Lnl1/cosxl = Lnl1l- Lnlcosxl (and because ln1=0) = -Lnlcosxl
yup ... I think that's how to do it..
sec(x) = 1/cos(x)
Lnlsec(x)l=-Lnlcos(x)l
Gives us:
I: e^(Lnlsec(x)l)
II: e^(-Lnlcos(x)l)
Next step:
I: e^(Lnlsec(x)l)=|sec(x)|=|1/cos(x)|=1/|cos(x)|
II: e^(-Lnlcos(x)l) = 1/e^(Lnlcos(x)l) = 1/|cos(x)|
I = II
Maybe something like that?
:o Omg is that all? seems so short.. D: Ahh wtheck can't work with those numbers.
[quote=KidZombie]Those arent numbers.[/quote]
You need to learn math.
Ln means natural log (log base e), while lxl means absolute value of x.