Cosh(2z) Equals Cosh^2(z) Plus Sinh^2(z)

[Wardlaw]

Show that cosh(2z)=cosh^2(z)+sinh^2(z)

?

 

[sjb-2812]

Show that cosh(2z)=cosh^2(z)+sinh^2(z)

?

Do you know Osborn's rule? ( http://en.wikipedia.org/wiki/Osborn's_Rule#Similarities_to_circular_trigonometric_functions )

 

[Wardlaw]

Do you know Osborn's rule? ( http://en.wikipedia.org/wiki/Osborn's_Rule#Similarities_to_circular_trigonometric_functions )

Yeah. I tried using the standard form for these expressions, when considering the RHS. I am then left with a quarter e^2z. Could you check this please?

 

[sjb-2812]

Hmm, I was just considering z as a random variable label, could just as easily be a, theta, or x.

So comparing to cos(2z) == cos²z - sin²z, there is a product of 2 sines, which you flip the sign of when comparing to hyperbolics, so cosh(2z) == cosh²z + sinh²z

 

[tiny-tim]

Welcome to PF!

Hi Wardlaw!

(try using the X² tag just above the Reply box )

Show that cosh(2z)=cosh^2(z)+sinh^2(z)

?

Yeah. I tried using the standard form for these expressions, when considering the RHS. I am then left with a quarter e^2z. Could you check this please?

You should get some e^-2z also.

Show us what you got for the RHS. ​

 

[Wardlaw]

Hi Wardlaw!

(try using the X² tag just above the Reply box )



You should get some e^-2z also.

Show us what you got for the RHS. ​

Oh yeah you are correct, my mistake. I can't even read my own working :)
How exactly do you go about solving thi problem?

 

[tiny-tim]

How exactly do you go about solving thi problem?

I leave it to you.

 

[Wardlaw]

I leave it to you.

Solved

 

[tiny-tim]

Woohoo! ​