Eliminate the parameter question

Question

abbles · Answer

x = sinht, y = cosht

abbles · Answer

Here's what I did:
x^2 = sin^ht 
y^2 = cos^2ht
x^2 + y^2 = cos^ht + sin^2ht
x^2 + y^2 = 1

HOWEVER, the answer book says the answer is y^2 - x^2 = 1
How are they getting a negative x??

irishboy123 · Answer

cos the hyperbolic identity is 

$cosh^2 t - sin^2 t = 1$

abbles · Answer

I thought the identity was $$\cos^2t + \sin^2t = 1$$

abbles · Answer

When were those hyperbolic identities supposed to be taught?  Precalculus?

solomonzelman · Answer

Well, in case you might forget the identity, you can derive it.
(Although I would recommend to rather remember them first, if you can.)

$\color{black}{\displaystyle \sinh t=\frac{e^t-e^{-t}}{2}}$   and   $\color{black}{\displaystyle \cosh t=\frac{e^t+e^{-t}}{2}}$

Then,
$\color{black}{\displaystyle \sinh^2t+\cosh^2t=\frac{(e^t-e^{-t})^2+(e^t+e^{-t})^2}{4}}$$\[1.8em]$
$\color{black}{\displaystyle \sinh^2t+\cosh^2t=\frac{(e^{2t}-2e^{t}e^{-t}+e^{-2t})+(e^{2t}+2e^{t}e^{-t}+e^{-2t})}{4}}$$\[1.8em]$
$\color{black}{\displaystyle \sinh^2t+\cosh^2t=\frac{(2e^{2t}+2e^{-2t})}{4}=\cosh 2t}$$\[1.8em]$

and
$\color{black}{\displaystyle \sinh^2t-\cosh^2t=\frac{(e^t-e^{-t})^2-(e^t+e^{-t})^2}{4}}$$\[1.8em]$
$\color{black}{\displaystyle \sinh^2t-\cosh^2t=\frac{(e^{2t}-2e^{t}e^{-t}+e^{-2t})-(e^{2t}+2e^{t}e^{-t}+e^{-2t})}{4}}$$\[1.8em]$
$\color{black}{\displaystyle \sinh^2t-\cosh^2t=\frac{--4}{4}=-1}$$\[1.8em]$

solomonzelman · Answer

I meant to write one negative on top, not twice ...

solomonzelman · Answer

then, you know  $\color{black}{\displaystyle \cosh^2t-\sinh^2t=1}$

agent0smith · Answer

Nice job @SolomonZelman

solomonzelman · Answer

Thank you!
I haven't done those for a while now:)

abbles · Answer

Thanks!  When were the hyperbolic identities supposed to be learned?  I must have glazed over those somewhere...

solomonzelman · Answer

Well, the main ones you have to remember are $\[0.7em]$
$\color{black}{\displaystyle \cosh t=\frac{e^t+e^{-t}}{2}}$   and   $\color{black}{\displaystyle \sinh t=\frac{e^t-e^{-t}}{2}}$ $\[1.3em]$
(The remaining hyperbolics can be easily derived from those  ...
e.g.  tanh(t)=sinh(t)/cosh(t)  or  sech(t)=1/cosh(t) .. etc.)$\[1.3em]$
Or, verbally, cosh is the average between $\color{black}{\displaystyle e^t}$ and $\color{black}{\displaystyle e^{-t}}$,
not to confuse which is which.

solomonzelman · Answer

you may also note $\cosh x=\cos(ix)$ ... for example

abbles · Answer

Mm.  Okay, thanks !

agent0smith · Answer

You might have just never learned them. I think I learned them in my geometry and trigonometry class in year 11 (I don't remember for sure), but then most precalculus classes don't seem to include any mention of the hyperbolic trig functions (like when I taught precalc).