Prove the identity cosh(x+y)=coshxcoshy+sinhxsinhy.
I think I understand why it\'s a positive sign but I don\'t know how to show it algebraically...
Solution
coshx = (e^x + e^-x)/2, sinhx = (e^x - e^-x)/2, then
coshxcoshy + sinhxsinhy = (1/2)(e^x + e^-x)(1/2)(e^y + e^-y) + (1/2)(e^x - e^-x) (1/2)(e^y - e^-
y)
= (1/4)(e^x + e^-x)(e^y + e^-y) + (1/4)(e^x - e^-x) (e^y - e^-y)
= (1/4)[e^(x+y) + e^(x-y) + e^(-x+y) + e^(-x-y)] + (1/4)[e^(x+y) - e^(x-y) - e^(-x+y) + e^(-x-y)]
= (1/4)[2e^(x+y)+2e^(-x-y)]
= (1/2)[e^(x+y) + e^-(x+y)]
= cosh(x+y)..
![Prove the identity cosh(x+y)=coshxcoshy+sinhxsinhy.
I think I understand why it's a positive sign but I don't know how to show it algebraically...
Solution
coshx = (e^x + e^-x)/2, sinhx = (e^x - e^-x)/2, then
coshxcoshy + sinhxsinhy = (1/2)(e^x + e^-x)(1/2)(e^y + e^-y) + (1/2)(e^x - e^-x) (1/2)(e^y - e^-
y)
= (1/4)(e^x + e^-x)(e^y + e^-y) + (1/4)(e^x - e^-x) (e^y - e^-y)
= (1/4)[e^(x+y) + e^(x-y) + e^(-x+y) + e^(-x-y)] + (1/4)[e^(x+y) - e^(x-y) - e^(-x+y) + e^(-x-y)]
= (1/4)[2e^(x+y)+2e^(-x-y)]
= (1/2)[e^(x+y) + e^-(x+y)]
= cosh(x+y).](https://image.slidesharecdn.com/provetheidentitycoshxycoshxcoshysinhxsinhy-230326113952-2a163d0d/85/Prove-the-identity-cosh-x-y-coshxcoshy-sinhxsinhy-I-think-I-un-pdf-1-320.jpg)
