Trig identity to hyperbolic identity

By Anonymous on Monday, March 12, 2001 - 07:17 pm :

I am having trouble with this question.

From the identity sin(Z + W) = sinZcosW + cosZsinW
find the corresponding identity for sinh(Z + W). thanks.

By Kerwin Hui (Kwkh2) on Tuesday, March 13, 2001 - 02:03 pm :

Do you know anything about complex numbers? If so, use the result

sin (ix) = i sinh x
cos (ix) = cosh x

If not, please write back.

Kerwin

By Anonymous on Tuesday, March 13, 2001 - 02:09 pm :

I haven't done a lot of complex number work, but have seen the result you have stated. However, I am still not sure how these results can be used to show what we are looking for. Your help is appreciated.

By Kerwin Hui (Kwkh2) on Tuesday, March 13, 2001 - 04:13 pm :

OK, a way without using complex numbers is to use the definition of sinh and cosh to prove that

$\sinh (A + B) \equiv \sinh A \cosh B + \cosh A \sinh B$

Kerwin

By Anonymous on Wednesday, March 14, 2001 - 01:10 am :

thanks

By The Editor :

A bit more elaboration on the two methods above:

sinh(Z+W)=(1/i)sin(iZ+iW) from Kerwin's identities
Expand this using the sin identity, to get seomthing in terms of cos(iZ), cos(iW), sin(iZ), etc.
Now use Kerwin's identities to turn those into coshZ, coshW, sinh(Z), etc.
Use the definitions sinhx=(e^x -e^-x )/2, coshx=(e^x +e^-x )/2 on the right-hand side of the identity, and it should simplify to what you need.