# Hyperbolic Functions

• Nov 19th 2008, 11:09 AM
Lonehwolf
Hyperbolic Functions
Well, I was a lousy ass during the lessons so I ended up quite rundown by a truck when I checked out the questions.. private lessons turned out to help a great deal, but i'm stuck with this one question that i really can't move past it (i skipped it for the time being but would like getting it solved for once :P)

Solve, for real values of x, the following equation:

$\displaystyle 4 tanh^2 x - sech x = 1$

Trying to open it up using the sinh x and cosh x rules got me a little bit lost in it

$\displaystyle sinh x = \frac{ e^x - e^{-x}}{2}$

and $\displaystyle cosh x = \frac{e^x + e^{-x}}{2}$

Solution to that one would be awesome :)

One more question my brain cells just won't go past is this one:

Solve, for real values of x, the following equation:
$\displaystyle sinh^2 x - 5 cosh x + 5 = 0$

The powers just confuse me, trying to use the identities led me to

$\displaystyle (cosh 2x -1) - 10 cosh x + 9 = 0$

(...messing a bit around with it...)

$\displaystyle e^{2x} + e^{-2x} - 10e^x + 10e^{-x} + 18 = 0$

And from then onwards, my brain shuts down. I can't get past that =/
• Nov 19th 2008, 11:12 AM
Lonehwolf
Just noticed I should have used the sin / cos identities for killing the power first... but I'd still get stuck anyways >_<
• Nov 19th 2008, 03:43 PM
Soroban
Hello, Lonehwolf!

I used the identity: .$\displaystyle \text{sech}^2x + \text{tanh}^2x \:=\:1$

Quote:

Solve, for real $\displaystyle x\!:\;\;4\:\!\text{tanh}^2x - \text{sech }\!x \:= \:1$

We have: .$\displaystyle 4(1 - \text{sech}^2x) - \text{sech }x \:=\:1 \quad\Rightarrow\quad 4\:\!\text{sech}^2x + \text{sech }x - 3 \:=\:0$

Factor: .$\displaystyle (\text{sech }\!x + 1)(4\:\!\text{sech }\!x -3)\:=\:0$

Then: .$\displaystyle \text{sech }\!x + 1 \:=\:0 \quad\Rightarrow\quad \text{sech }\!x \:=\:-1$ . . . Impossible: $\displaystyle \text{sech }\!x$ is always positive.

And: .$\displaystyle 4\:\!\text{sech }\!x - 3 \:=\:0 \quad\Rightarrow\quad \text{sech }\!x \:=\:\frac{3}{4} \quad\Rightarrow\quad \cosh x \:=\:\frac{4}{3}$

We have: .$\displaystyle \frac{e^x + e^{-x}}{2} \:=\:\frac{4}{3} \quad\Rightarrow\quad 3e^x + 3e^{-x} \:=\:8 \quad\Rightarrow\quad 3e^{2x} -8e^x + 3 \:=\:0$

Quadratic Formula: .$\displaystyle e^x \;=\;\frac{4\pm\sqrt{7}}{3}$

Therefore: .$\displaystyle x \;=\;\ln\left(\frac{4\pm\sqrt{7}}{3}\right) \;\approx\; \begin{Bmatrix}\;0.795365461 \\ \text{-}0.795365461 \end{Bmatrix}$

• Nov 20th 2008, 02:01 AM
Lonehwolf
You're something =o

That makes definitely a lot of sense to me, so 1 problem outta 2 is solved (Punch)

Regarding the second problem, I still can't get it to budge :( Any help much appreciated. I'll give it another try once i finish the last exercise.

$\displaystyle sinh^2 x - 5 cosh x + 5 = 0$
• Nov 20th 2008, 04:46 AM
Soroban
Hello again, Lonehwolf!

Quote:

Solve: .$\displaystyle \sinh^2\!x - 5\cosh x + 5 \:=\: 0$
Identity: .$\displaystyle \cosh^2\!x - \sinh^2\!x \:=\:1$

We have: .$\displaystyle \sinh^2\!x - 5\cosh x + 5 \:=\:0$

Then: .$\displaystyle \overbrace{(\cosh^2\!x - 1)} - 5\cosh x + 5 \:=\:0 \quad\Rightarrow\quad \cosh^2\!x - 5\cosh x + 4 \:=\:0$

Factor: .$\displaystyle (\cosh x - 1)(\cosh x - 4)\:=\:0$

Hence: .$\displaystyle \begin{Bmatrix}\cosh x \:=\:1 & \Rightarrow & \dfrac{e^x+e^{-x}}{2} \:=\:1 \\ \\[-3mm] \cosh x \:=\:4 & \Rightarrow & \dfrac{e^x+e^{-x}}{2} \:=\:4 \end{Bmatrix} \quad\hdots\quad\text{etc.}$