baldymcgee6 Group Title Show that sinh(3t)*sinh(t)=1+2cosh(2t) for t≠0 one year ago one year ago

1. baldymcgee6 Group Title

@LolWolf, @AccessDenied, @lgbasallote

2. baldymcgee6 Group Title

wait, thats wrong

3. baldymcgee6 Group Title

sinh(3t)/sinh(t)=1+2cosh(2t) for t≠0

4. Algebraic! Group Title

they aren't equal...try again

5. Algebraic! Group Title

ok finally

6. baldymcgee6 Group Title

@Algebraic! do you know how to do it?

7. Algebraic! Group Title

yep

8. AccessDenied Group Title

$\text{sinh} \; x = \frac{e^x - e^{-x}}{2}$ 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.

9. AccessDenied Group Title

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.

10. baldymcgee6 Group Title

@AccessDenied YOU ROCK, thanks so much!!

11. AccessDenied Group Title

I should note that I am using x instead of t. My bad. :P

12. AccessDenied Group Title

and, I'm glad to help! :)

13. baldymcgee6 Group Title

thanks so much again

14. Algebraic! Group Title

don't forget, you can redeem your medals for cash prizes at the end of every month.

15. LolWolf Group Title

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

16. baldymcgee6 Group Title

lol, thanks for you valiant effort @LolWolf

17. LolWolf Group Title

Valiantly late, haha, but, yes