# hyperbolic function identity

• Feb 24th 2007, 02:48 AM
gracy
hyperbolic function identity
By writing JA for in cot^2 +1=cosec^2 determine the corresponding hyperbolic identify
• Feb 24th 2007, 05:48 AM
CaptainBlack
Quote:

Originally Posted by gracy
By writing JA for in cot^2 +1=cosec^2 determine the corresponding hyperbolic identify

The trick you need here is to change every term containing sin^2 to -sinh^2
(after converting it so it contains sin's and cos's only) in a trig identity to get
the corresponding hyperbolic identity.

So

cot^2 +1=cosec^2

becomes:

sin^2/cos^2 + 1 = 1/sin^2.

So the corresponding hyperbolic identity is:

-sinh^2 / cosh^2 + 1 = -1/sinh^2

or:

-tanh^2 + 1 = -cosech^2

or:

tanh^2 -1 = cosech^2

RonL
• Feb 24th 2007, 12:32 PM
Soroban
Hello, gracy!

Two questions: What is "JA"? .And do they really expect us to know
. . conversions between circular and hyperbolic trig?

Quote:

By writing in JA for: cot²x + 1 .= .csc²x
determine the corresponding hyperbolic identify.

I would begin with the first identity of hyperbolics: .cosh²x - sinh²x .= .1

. . . . . . . . . . . . . .cosh²x . .sinh²x . . . . . .1
Divide by sinh²x: . --------- - -------- .= . --------
. . . . . . . . . . . . . .sinh²x . . sinh²x . . . .sinh²x

. . and we get: .coth²x - 1 .= .csch²x