What is the hyperbolic function of

(cosh A *cosh B)^2

obviously it is the same thing as cosh^2(A) * cosh^2(B).

Is there any way I could get it to become

cosh^2(A) * cosh^2(B) = Cosh^2(A) + Cosh^2(B) ?

Printable View

- Nov 29th 2010, 03:11 AMTurloughmackHyperbolic Functions
What is the hyperbolic function of

(cosh A *cosh B)^2

obviously it is the same thing as cosh^2(A) * cosh^2(B).

Is there any way I could get it to become

cosh^2(A) * cosh^2(B) = Cosh^2(A) + Cosh^2(B) ? - Nov 29th 2010, 03:12 AMmr fantastic
- Nov 29th 2010, 03:13 AMdwsmith
Try using $\displaystyle \displaystyle \left(\frac{e^a-e^{-a}}{2}\right)^2*\left(\frac{e^b-e^{-b}}{2}\right)^2$ and I don't know if it will work but it is worth a start.

- Nov 29th 2010, 03:17 AMTurloughmack
From the cosine law, we can write sin^2(gamma) = 1 - (AB - C / sinh a.sinh b)^2 [where A=cosh a, B= cosh b, C = cosh c]

I have done this. The next part of the question says

deduce

sin^2(gamma) sinh^2(a) sinh^2(b) = 1 - A^2 - B^2 - C^2 +2ABC.

This is where my problem lies, I understand how to get the LHS. I dont see hot to get the RHS - Nov 29th 2010, 03:21 AMdwsmith
Start using the hyperbolic identities to see what you can do.

Hyperbolic function - Wikipedia, the free encyclopedia