Hi , I'm stuck on part b) of the attached proving cosh(x-ln2)=sinh(x) .
Substituting and simplifying seems to give me e^(x)-e^(-x) and not with the half to prove the identity .
Hi , I'm stuck on part b) of the attached proving cosh(x-ln2)=sinh(x) .
Substituting and simplifying seems to give me e^(x)-e^(-x) and not with the half to prove the identity .
I really do not follow what is going on in that attachment.
By here are two facts.
$\displaystyle e^{x-\ln(2)}=\frac{e^x}{2}$ so $\displaystyle \cosh(x-\ln(2))=\frac{e^{2x}+4}{4e^x}$