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baldymcgee6
Show that sinh(3t)*sinh(t)=1+2cosh(2t) for t≠0
@LolWolf, @AccessDenied, @lgbasallote
sinh(3t)/sinh(t)=1+2cosh(2t) for t≠0
they aren't equal...try again
@Algebraic! do you know how to do it?
\[ \text{sinh} \; x = \frac{e^x - e^{-x}}{2} \] We can rewrite the left-hand side to match the right-hand 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.
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.
@AccessDenied YOU ROCK, thanks so much!!
I should note that I am using x instead of t. My bad. :P
and, I'm glad to help! :)
thanks so much again
don't forget, you can redeem your medals for cash prizes at the end of every month.
So we know: \[ \sinh x=\frac{e^x-e^{-x}}{2} \]Therefore: \[ \sinh 3x=\frac{e^{3x}-e^{-3x}}{2} \]So: \[ \frac{\sinh(3t)}{\sinh(t)}=\frac{2}{e^x-e^{-x}}\cdot\frac{e^{3x}-e^{-3x}}{2}=\frac{2e^x}{e^{2x}-1}\cdot\frac{e^{6x}-1}{2e^{3x}}=\\ \frac{2e^x}{e^{2x}-1}\cdot\frac{e^{6x}-1}{2e^{3x}}=\frac{(e^{2x}-1)(e^{4x}+e^{2x}+1)}{e^{2x}(e^{2x}-1)}=\\ \frac{e^{4x}+e^{2x}+1}{e^{2x}}=e^{2x}+1+e^{-2x}=1+2\cosh x \]Ahh, this takes forever... and I mad a mistake halfway through, so I had to restart... anyways, +1 internets to @AccessDenied
lol, thanks for you valiant effort @LolWolf
Valiantly late, haha, but, yes