How to solve a hyperbolic (cosh) equality when the argument of the cosh function is different

Question

if
$ \alpha+\lambda = c \cosh( \frac{a+d}{c})$ , and
$ \alpha + \lambda = c \cosh( \frac{-a+d}{c})$

how does one get that $α + λ = c \cosh(\frac{a}{c})$ and $d = 0$

is there a rule that im missing?

d is a constant

edit:

this is what i did: since cosh is an even function and is bijective, $\cosh(u)= \cosh(v) \implies u=c$ so $ \frac{a+d}{c} = \frac{-a+d}{c} \implies -a+d = a+d \implies a = 0$ did i make a mistake ? i cant seem to get $d=0$

Answer 1

You want to know under what condition $$ \cosh(\frac{a+b}{c})=\cosh(\frac{-a+b}{c}) $$ Expanding the sum of arguments for $\cosh$, $$ \cosh(\frac{a+b}{c})=\cosh(\frac{a}{c})\cosh(\frac{b}{c})+\sinh(\frac{a}{c})\sinh(\frac{b}{c}) $$ and observe that $\cosh$ is even and $\sinh$ is odd $$ \cosh(\frac{-a+b}{c})=\cosh(\frac{-a}{c})\cosh(\frac{b}{c})+\sinh(\frac{-a}{c})\sinh(\frac{b}{c})\\ =\cosh(\frac{a}{c})\cosh(\frac{b}{c})-\sinh(\frac{a}{c})\sinh(\frac{b}{c}) $$ Thus the equality can only hold if $a=0$ or $b=0$ or both $a$ and $b$ equal zero.

Answer 2

Observe $\cosh$ is increasing (hence one-to-one) on $[0,\infty)$, and similarly is decreasing (hence one-to-one) on the interval $(-\infty,0]$. Moreover, $\cosh$ is even (symmetric across the $y$-axis).

Therefore, $\cosh u=\cosh v$ implies $v=\pm u$. So the system gives either

• $a=0$ and $d\ne0$ and $\alpha+\lambda = c\cosh(d/c)$, or
• $a\ne0$ and $d=0$ and $\alpha+\lambda=c\cosh(a/c)$

depending on the sign of $\frac{-a+d}{c}=\pm\frac{a+d}{c}$.