How do you show that #cosh^2 x - sinh^2 x = 1# ?
1 Answer
Oct 15, 2017
See explanation...
Explanation:
#cosh x = 1/2(e^x+e^(-x))#
#sinh x = 1/2(e^x-e^(-x))#
So:
#cosh^2 x - sinh^2 x = 1/4((e^x+e^(-x))^2-(e^x-e^(-x))^2)#
#color(white)(cosh^2 x - sinh^2 x) = 1/4((e^(2x)+2+e^(-2x))-(e^(2x)-2+e^(-2x)))#
#color(white)(cosh^2 x - sinh^2 x) = 1#
