Question

cosh2x = cosh^2x + sinh^2x

Prove this identity!

Answer 1

\[\cosh2x = \cosh^2x + \sinh^2x\]

Answer 2

cosh^2x, do you mean cos (h^2x) ?

Answer 3

no this is hyperbolic functions lol

Answer 4

Calc 2, wio knows what's up

Answer 5

Dang, ....

Answer 6

Okay, so they probably don't want you to use Osborn's rule here.

Answer 7

nope lol

Answer 8

wio - calculator.

Answer 9

I know that

\[\cosh^2x = (\frac{ e^x+e^-x }{ 2 })^2\]

and

\[\sinh^2x = (\frac{ e^x-e^-x }{ 2 })^2\]

Answer 10

-x should be ^

Answer 11

So I think we should make cosh^2x on the left side basically = \[(e^x+e^-x/2)^2 \]

Answer 12

What ya think wio?

Answer 13

It's simple. Expand out the left hand side. Expand out the right hand side.

Answer 14

All the expanding you do on the right hand side can be done in reverse, as if you were really good at factoring, for the actual proof.

Answer 15

i always forget cosh^2x = (coshx)(coshx)

Answer 16

haha

Answer 17

When you expand \(\cosh^2(x)\) notice:\[

\cosh^2(x) = \frac{e^{2x}+2e^0+e^{-2x}}{4}
\]

Answer 18

yup yup

Answer 19

Just wrote that down as well

Answer 20

So when doing things in reverse, you have to subtract 2 and add 2 at the same time.

Answer 21

wio is the best! I told you!
Dr calculator....!

Answer 22

@wio is e^x * -e^-x = -1?

Answer 23

e=2.7

Answer 24

right

Answer 25

Yeah, that is correct.

Answer 26

kk

Answer 27

What exactly does cosh2x equal to?

Answer 28

is it just e^2x+e^-2x/2 ?

Answer 29

Yes.

Answer 30

mhm ok, let me see if I got it now

Answer 31

nope..

Answer 32

@wio I did something wrong, I'm getting \[e^{2x} + e^{-2x} + 2\]

Answer 33

You can rewrite the right side as \(2sinh^2x +1\) and save yourself a lot of stupid multiplying

Answer 34

I really wish you mentioned that before lmao

Answer 35

I'll give it a shot and see what I get this time >.<

Answer 36

Well what @SACAPUNTAS said is true, but they might not want that depending on the rigor involved.

Answer 37

This is why I hate identities.

Answer 38

\[\begin{split}
\cosh^2(x) + \sinh^2(x)
&=\left(\frac{e^{x}+e^{-x}}{2}\right)^2+\left(\frac{e^{x}-e^{-x}}{2}\right)^2\\
&=\frac{e^{2x}+2+e^{-2x}}{4}+\frac{e^{2x}-2+e^{-2x}}{4}\\
&=\frac{e^{2x}+e^{2x}+\cancel{2-2}+e^{-2x}+e^{-2x}}{4}\\
&=\frac{2e^{2x}+2e^{-2x}}{4}\\
&=\frac{e^{2x}+e^{-2x}}{2}\\
&=\cosh(2x)

\end{split}
\]

Answer 39

Write this backwards.

Answer 40

Ah very very nice, thank you so much Wio!!!

Answer 41

I found my mistake as well haha, I added the 2s >_>