Hyperbolic Functions

[furrybean]

Anyone here any good at hyperbolic functions and complex number?

Ive got a question that I'm struggling on...

Given that sinjA = jsinhA, show that the hyperbolic identity corresponding to sin3θ = 3sinθ - 4sin3θ is: sinh3A = 3sinh A + 4sinh3A by writing jA for θ

:shock:

[furrybean]

](*,) ](*,) :help: :help: #-o

[cuffy]

In mathematics, the hyperbolic functions are analogs of the ordinary trigonometric, or circular, functions. The basic hyperbolic functions are the hyperbolic sine "sinh", and the hyperbolic cosine "cosh", from which are derived the hyperbolic tangent "tanh", etc., in analogy to the derived trigonometric functions. The inverse functions are the inverse hyperbolic sine "arsinh" (also called "arсsinh" or "asinh") and so on.

HTH
:D

[furrybean]

Close but no cigar...

Thanks for the help but it needs to be rearranged using those identities somehow.

Cheers anyway

[wyrdness]

Try PMing Professor. He's a Professor of Mathematics.

[Fearg]

Anyone here any good at hyperbolic functions and complex number?

Ive got a question that I'm struggling on...

Given that sinjA = jsinhA, show that the hyperbolic identity corresponding to sin3θ = 3sinθ - 4sin3θ is: sinh3A = 3sinh A + 4sinh3A by writing jA for θ

:shock:

Is that question right? should it be sin3θ = 3sinθ - 4sin^3θ (i.e. 4 sin cubed theta)?

[furrybean]

yup, thats what I got in front of me. I got the answer except I cant get the -4sin3θ to + 4sin3θ

[Fearg]

Anyone here any good at hyperbolic functions and complex number?

Ive got a question that I'm struggling on...

Given that sinjA = jsinhA, show that the hyperbolic identity corresponding to sin3θ = 3sinθ - 4sin3θ is: sinh3A = 3sinh A + 4sinh3A by writing jA for θ

:shock:

sin3θ = 3sinθ - 4sin^3θ

substituting jA for θ gives

sin3jA = 3sinjA - 4sin^3jA

since sinjA = jsinhA

jsinh3A = 3jsinhA - 4j^3sinh^3A

divding thorough by j gives

sinh3A = 3sinhA - 4j^2sinh^3A

but as j^2 = -1

sinh3A=3sinhA+4sinh^3A

[Jabba]

sin3θ = 3sinθ - 4sin^3θ

substituting jA for θ gives

sin3jA = 3sinjA - 4sin^3jA

since sinjA = jsinhA

jsinh3A = 3jsinhA - 4j^3sinh^3A

divding thorough by j gives

sinh3A = 3sinhA - 4j^2sinh^3A

but as j^2 = -1

sinh3A=3sinhA+4sinh^3A

Geek alert :shock:

[Fearg]

sin3θ = 3sinθ - 4sin^3θ

substituting jA for θ gives

sin3jA = 3sinjA - 4sin^3jA

since sinjA = jsinhA

jsinh3A = 3jsinhA - 4j^3sinh^3A

divding thorough by j gives

sinh3A = 3sinhA - 4j^2sinh^3A

but as j^2 = -1

sinh3A=3sinhA+4sinh^3A

Geek alert :shock:

:oops: :oops: :oops:

[Jabba]

:oops: :oops: :oops:

:smt056

[furrybean]

It aint cubed on the assignment sheet but do u reckon it might me wrong.

I got to this

sin3θ = 3sinθ - 4sin3θ

substituting jA for θ gives

sin3jA = 3sinjA - 4(sin3jA)

since sinjA = jsinhA

jsinh3A = 3jsinhA - 4(jsinh3A)

jsinh3A = 3jsinhA - 4jsinh3A

just that negative sign!!!

[Fearg]

It aint cubed on the assignment sheet but do u reckon it might me wrong.

I got to this

sin3θ = 3sinθ - 4sin3θ

substituting jA for θ gives

sin3jA = 3sinjA - 4(sin3jA)

since sinjA = jsinhA

jsinh3A = 3jsinhA - 4(jsinh3A)

jsinh3A = 3jsinhA - 4jsinh3A

just that negative sign!!!

Personally yes I think the assignment sheet is wrong, won't be the first time! or even the last! :lol:
Don't sweat over it anymore

[furrybean]

well cheers for the help. Can you go that with the 4sin3jA and put the 4 like 4(sin3jA) so it goes to 4jsinh3A

[Fearg]

well cheers for the help. Can you go that with the 4sin3jA and put the 4 like 4(sin3jA) so it goes to 4jsinh3A

I got to this

sin3θ = 3sinθ - 4sin3θ

substituting jA for θ gives

sin3jA = 3sinjA - 4sin3jA

since sinjA = jsinhA

jsinh3A = 3jsinhA - 4jsinh3A

just that negative sign!!!

I would leave the brackets out personally (as above)

then divide thru by j

sinh3A = 3sinhA - 4sinh3A

but like you say that's not what is on your sheet

[furrybean]

Ok well cheers for the help matey :thumbsup:

[Fearg]

The proof is in the text book, which have just dug out and blown the cobwebs off :D



http://upload5.postimage.org/241045/Image1.jpg (http://upload5.postimage.org/241045/photo_hosting.html)

[furrybean]

Minter, well I now know they dont make up the qu themselves.

What book is that?

Cheers

[Fearg]

Minter, well I now know they dont make up the qu themselves.

What book is that?

Cheers

Higher Engineering Mathematics by John Bird

what are you studying? :study:

[Rob S (Yella)]

The answer is

42

[furrybean]

Minter, well I now know they dont make up the qu themselves.

What book is that?

Cheers

Higher Engineering Mathematics by John Bird

what are you studying? :study:

Electrical Engineering. What were you studying?

[Fearg]

Electrical Engineering. What were you studying?

Mechanical Engineering #-o

[thor]

I did that stuff when I was 17. Shame I've forgotten all of it... :(

[Alpinestarhero]

Mmmm maths, highly sexy subject.

Not good with hyperbolic functions; remind me what they look like on a graph though? Maybe I can have a go after my exams are finished (when does it need to be done by, and whats it for?).

Matt

[Alpinestarhero]

sin3θ = 3sinθ - 4sin^3θ

substituting jA for θ gives

sin3jA = 3sinjA - 4sin^3jA

since sinjA = jsinhA

jsinh3A = 3jsinhA - 4j^3sinh^3A

divding thorough by j gives

sinh3A = 3sinhA - 4j^2sinh^3A

but as j^2 = -1

sinh3A=3sinhA+4sinh^3A

Geek alert :shock:

Maths is a wonderful subject, masters of it should be worshiped as gods :D

Matt

[furrybean]

BTW it WAS cubed!!!!

[Fearg]

:smt023 :smt023 :smt023