Question

Hyperbolic functions:

Answer 1

http://prntscr.com/32y8io

Answer 2

omg

Answer 3

hard to eyeball all of them but just as cosine is even, so is cosh, and C is the proof

Answer 4

For some of these, they play some very subtle tricks on you...

Answer 5

think the first one looks good too
\[\cosh(x)+\sinh(x)=e^x\] for sure

Answer 6

Yea, the cosh(-x) was the first one I knew could be right.

Answer 7

first is right as well

Answer 8

like for (c)
cosh(x) = 1/2 (e^x + e^(-(-x)) ) = 1/2 (e^ {-x} \(\color{red}-\) e^{x} ) where does that negative sign come from?

Answer 9

cosh(-x) = 1/2 e^(-x) <-- i wrote that part incorrectly

Answer 10

I think for that one they factored out a -1?

Answer 11

It does have a sign change

Answer 12

Oh wait, nvm I looked at the wrong side

Answer 13

well, the only negative signs are in the powers. you can't really convert the power into a multiplication of -1. And to factor out that -1, it would be coming from both e^(-x) and e^(x). ;x

Answer 14

oh doh i wasn't paying attention
it says "which one is PROVED correctly" not "which identity is right

Answer 15

first one is proved correctly for sure (i think)

Answer 16

We never did get see cosh(-x)=cosh proved so I'm not sure if that one is right >_<

Answer 17

i agree, first one looks good to me.
cosh (-x) = cosh (x) is almost true, except instead of that negative sign:
\( \cosh (-x) = \dfrac{1}{2} (e^{-x} + e^{-(-x)} ) = \dfrac{1}{2} (e^{-x} + e^{+x} ) \)
that's just a matter of commutative property now,

Answer 18

it is true, the proof is wrong
replace \(x\) by \(-x\) in
\[\frac{1}{2}\left(e^x+e^{-x}\right)\] and you get it right away

Answer 19

Alright so first off, do you even KNOW what cosh(t) and sinh(t) are?

They satisfy the following equation on the unit hyperbola,

\[x^2-y^2=1\]

where

x=cosh(t)
y=sinh(t)

So when x=1 then obviously we can see on the graph above that y=0 since x^2-y^2=1.

Also note that the cosh(t) (t is just the hyperbolic angle) will give you the same answer if we travel to positive or negative amounts since you will always have something 1 or greater while since sinh(t) tells you the y-part of the hyperbola you can easily see how it'll be positive on the top and negative on the bottom.

They're really similar to the trig functions for the circle. Think about them. Play with them. Also note that they are both their own second derivatives, not their own negative second derivatives like the circular trig functions are, which is interesting.