If sinhx = (e^x - e^x)/2 & coshx = (e^x + e^-x)/2
I think you're supposed to substitute the equivalents of sinhx and coshx into the orig. equation. I just think I'm messing up in the algebra.
2007-10-21 07:04:53 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
sorry I messed up the original, it's coshx^2 - sinhx^2 = 1
2007-10-21 07:13:53 · update #1
does anybody realize I already substituted into the equation? I just need a clarification on my algebra THANK YOU!
2007-10-21 07:33:37 · update #2
You're essentially trying to prove
(e^x - e^x)/2 - (e^x + e^-x)/2 = 1
I think you'll have to use series expansion like Taylor or McLaurin series expansion to show this if I remember correctly.
2007-10-21 07:24:58 · answer #1 · answered by Shh! Be vewy, vewy quiet 6 · 0⤊ 0⤋
You don't because coshx^2 - sinhx^2 = 1
2007-10-21 14:12:41 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
Is not true
(cosh x)^2-(sinh x)^2=1
(e^2x+e^-2x+2 -e^2x-e^-2x+2)/4=1
2007-10-21 14:16:12 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋