Question

prove that cosh(2x) = 2sinh^2(x) + 1
*Recall that sinh(x) = (e^x - e^-x)/2 and
cosh(x) = (e^x + e^-x)/2...
Also, cosh^2(x) - sinh^2(x)=1
Can anyone help me?

Answer 1

substitute and prove LHS = RHS

Answer 2

\[2(\frac{e^x-e^{-x}}{2})^2+1=(\frac{e^{2x}-2+e^{-2x}}{2})+1=\]
make it one fraction by adding 1

Answer 3

Aani, I know that I am suppose to prove that the left equals the right...that is what a proof means...anything else you can contribute? Thanks

Answer 4

Hi Jonask...let me take a second and read over your post...thanks

Answer 5

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

Answer 6

@SinginDaCalc2Blues so first we put wat sinh is in the right hand side equation

Answer 7

HI again Jonask...I am trying to read your equations but I think the computer changed their format and I simply can't understand them...do you see what I see?

Answer 8

okay i'll send a picture of them

Answer 9

thanks!

Answer 10

I have tried solving down both sides many times...it looks like the right side might be better to manipulate but I'm not certain...what do you think?

Answer 11

Hey Jonask...I got it!!! Thank you! I was able to figure out the jibberish equations! Awesome! Thank u!

Answer 12

here you go took some time

Answer 13

yeah i thought i had it but i still had a 2 in front...one second and i will look at your work...

Answer 14

the two went out because the denominator was 2^2=4 so canceling the two on top

Answer 15

Amazing work!...That's what I was missing, the darn 4 in the bottom! haha
Thank yiou soooo much for youe effort!!!
I would give yoiu 2 gold medals if I could!

Answer 16

now some other easy way
lets consider the right hand side again\[2\sinh^2x+1=2\sinh^2x+\cosh^2x-\sinh^2x=\sinh^2x+\cosh^2x=\cosh2x\]

Answer 17

@SinginDaCalc2Blues

Answer 18

going through your 2nd solution now...

Answer 19

I dont follow this last part, if I'm reading it correctly...
\sinh^2x+\cosh^2x=\cosh2x\]
Is that saying: (sinh^2x + cosh^2x) = cosh^2x ???
If I'm reading that correctly, I don't understand

Answer 20

I gave you a medal and became your fan!
If you happen to write anything else, I will come back here and check it out...thanks again!

Answer 21

Hey Jonask, if you feel like helping me one more time, I posted a new question...if you have the time

Answer 22

\[\sinh^2x+\cosh^2x=(\frac{e^x-e^{-x}}{2})^2+(\frac{e^x+e^{-x}}{2})^2=\cosh 2x \]
i notice that for the last part yuo have to show this 1st