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baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
@LolWolf, @AccessDenied, @lgbasallote
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
wait, thats wrong
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
sinh(3t)/sinh(t)=1+2cosh(2t) for t≠0
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
they aren't equal...try again
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
ok finally
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
@Algebraic! do you know how to do it?
 2 years ago

AccessDenied Group TitleBest 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.
 2 years ago

AccessDenied Group TitleBest 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.
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
@AccessDenied YOU ROCK, thanks so much!!
 2 years ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
I should note that I am using x instead of t. My bad. :P
 2 years ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
and, I'm glad to help! :)
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
thanks so much again
 2 years ago

Algebraic! Group TitleBest 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.
 2 years ago

LolWolf Group TitleBest 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
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
lol, thanks for you valiant effort @LolWolf
 2 years ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Valiantly late, haha, but, yes
 2 years ago
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