Hyperbolic Functions - SV650.org

15-01-07, 08:14 PM

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

15-01-07, 08:42 PM

15-01-07, 08:46 PM

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

15-01-07, 08:47 PM

Close but no cigar...

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

Cheers anyway

15-01-07, 09:16 PM

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

15-01-07, 09:28 PM

Quote:

Originally Posted by 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

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

15-01-07, 09:31 PM

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

15-01-07, 09:36 PM

Quote:

Originally Posted by 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

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

15-01-07, 09:39 PM

Quote:

Originally Posted by 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

15-01-07, 09:41 PM

Quote:

Originally Posted by Jabba

Quote:

Originally Posted by 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

15-01-07, 08:14 PM	#1
furrybean Guest Posts: n/a	Hyperbolic Functions 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

15-01-07, 08:46 PM	#3
cuffy Guest Posts: n/a	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

15-01-07, 08:47 PM	#4
furrybean Guest Posts: n/a	Close but no cigar... Thanks for the help but it needs to be rearranged using those identities somehow. Cheers anyway

15-01-07, 09:16 PM	#5
wyrdness Guest Posts: n/a	Try PMing Professor. He's a Professor of Mathematics.

15-01-07, 09:31 PM	#7
furrybean Guest Posts: n/a	yup, thats what I got in front of me. I got the answer except I cant get the -4sin3θ to + 4sin3θ