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baldymcgee6Best ResponseYou've already chosen the best response.0
@LolWolf, @AccessDenied, @lgbasallote
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
sinh(3t)/sinh(t)=1+2cosh(2t) for t≠0
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
they aren't equal...try again
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
@Algebraic! do you know how to do it?
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.2
\[ \text{sinh} \; x = \frac{e^x  e^{x}}{2} \] We can rewrite the lefthand side to match the righthand side. \[ \frac{e^{3x}  e^{3x}}{2} \div \frac{e^x  e^{x}}{2} \\ \frac{e^{3x}  e^{3x}}{\cancel{2}} \times \frac{\cancel{2}}{e^x  e^{x}}\\ \frac{e^{3x}  e^{3x}}{e^x  e^{x}} \\ \frac{(e^x)^3  (e^{x})^3}{e^x  e^{x}} \] If we consider the numerator as a difference of cubes, we can factor it like this: \( a^3  b^3 = (a  b)(a^2 + ab + b^2)\). Notice that this creates a factor in the denominator that is also in the numerator.
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.2
When the factors cancel, this remains: \[ \frac{\cancel{(e^x  e^{x})}((e^x)^2 + e^x e^{x} + (e^{x})^2)}{\cancel{e^x  e^{x}}} \\ = e^{2x} + 1 + e^{2x} \] Which starts to look a lot like 2cosh 2x + 1, it should be simple manipulation to justify that from there.
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
@AccessDenied YOU ROCK, thanks so much!!
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.2
I should note that I am using x instead of t. My bad. :P
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.2
and, I'm glad to help! :)
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
thanks so much again
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
don't forget, you can redeem your medals for cash prizes at the end of every month.
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
So we know: \[ \sinh x=\frac{e^xe^{x}}{2} \]Therefore: \[ \sinh 3x=\frac{e^{3x}e^{3x}}{2} \]So: \[ \frac{\sinh(3t)}{\sinh(t)}=\frac{2}{e^xe^{x}}\cdot\frac{e^{3x}e^{3x}}{2}=\frac{2e^x}{e^{2x}1}\cdot\frac{e^{6x}1}{2e^{3x}}=\\ \frac{2e^x}{e^{2x}1}\cdot\frac{e^{6x}1}{2e^{3x}}=\frac{(e^{2x}1)(e^{4x}+e^{2x}+1)}{e^{2x}(e^{2x}1)}=\\ \frac{e^{4x}+e^{2x}+1}{e^{2x}}=e^{2x}+1+e^{2x}=1+2\cosh x \]Ahh, this takes forever... and I mad a mistake halfway through, so I had to restart... anyways, +1 internets to @AccessDenied
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
lol, thanks for you valiant effort @LolWolf
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
Valiantly late, haha, but, yes
 one year ago
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