Question

tanh(a+b)

Answer 1

http://upload.wikimedia.org/wikipedia/en/math/7/e/0/7e011437dfac58bafd05c2d7818fa419.png
Why?

Answer 2

I got to http://upload.wikimedia.org/wikipedia/en/math/b/c/5/bc5fa29e66aef02ca27cb2f1c37cf55e.png
and
http://upload.wikimedia.org/wikipedia/en/math/9/6/4/96438ddb87fbc852570541e8ecfb04e7.png
but after that? Is this just a basic division question I'm mucking up?

Answer 3

\[ \sin hx = \frac{e^x - e^{-x}}{2} \]
\[ \cos hx = \frac{e^x + e^{-x}}{2} \]

Answer 4

\[\tanh(x+y)=(e^ye^x)^2-1/(e^ye^x)^2+1\]

Answer 5

http://upload.wikimedia.org/wikipedia/en/math/c/7/0/c70b0281ad41831f4dabeeddec5c58eb.png

Answer 6

Same thing. I've got it- it was just a mundane division thing

Answer 7

\[(\sinh(x+y)/\cosh(x+y)) \times(coshxcoshy)^{-1}/(coshxcoshy)^{-1}\]

Answer 8

simplification of
\[\frac{\tanh x +\tanh y}{1 - \tanh x \tanh y} = \LARGE \frac{\frac{e^x-e^{-x}}{e^x+e^{-x}}+\frac{e^y-e^{-y}}{e^y+e^{-y}}}{1 - \frac{e^x-e^{-x}}{e^x+e^{-x}}*\frac{e^y-e^{-y}}{e^y+e^{-y}}}\]
Yields,
\[ \tanh (x+y) = \frac{e^{x+y} - e^{-(x+y)}}{e^{x+y} + e^{-(x+y)}} \]