Help me to remember $\operatorname{cosh}^{2}(y) -\operatorname{sinh}^{2}(y)=1$, some easy verification and deduction? I can faintly visualize some way of deducing this formula with exponential functions but forgot it. How do you remember it? Suppose you just forget whether it is plus-or-minus there, how do you find the right formula now? Use pythagoras and?
 A: One approach is to remember that $\cosh x = \cos(ix)$ and $\sinh x = -i \sin (ix)$, which reduces this to $\sin^2 z + \cos^2 z = 1$. 
A: Maybe this will help as a mnemonic. 
The functions $\cosh$ and $\sinh$ are to the (right half of) the hyperbola $x^2-y^2=1$ as $\cos$ and $\sin$ are to the circle $x^2+y^2=1$. Or, more sloppily, $\cosh$ ($\cos$) is $x$ and $\sinh$ ($\sin$) is $y$.
By that I mean that the standard trigonometric parametrization of the unit circle is
$$x=\cos t\qquad y=\sin t.$$
A standard parametrization of the right half of the rectangular hyperbola $x^2-y^2=1$ is 
$$x=\cosh t\qquad y=\sinh t.$$
Remark: I like $\cos^2\theta+\sin^2\theta=1$ better than $\sin^2\theta+\cos^2\theta=1$. After all, $x$ comes before $y$.
A: The derivation is of the form (see that this checks out yourself)
$$\left(\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2=ab.$$
If $ab=1$ then we get the formula, specifically when $a=e^x$ and $b=e^{-x}$ and then the squares above are the two hyperbolic trig functions. This is just plugging in definitions and computing, that's all.
A: remember $(a+b)^2-(a-b)^2=4ab$ and $e^xe^{-x}=1$ 
A: If you’re visually oriented, try remembering the shapes of the graphs for $x\ge 0$, together with that of $y=e^x$: they’re very similar, they get closer together as $x$ increases, and $e^x$ is squeezed in above $\operatorname{sinh}x$ and below $\operatorname{cosh}x$. Since they’re blowing up, $\operatorname{cosh}^2 x+\operatorname{sinh}^2 x$ can’t be constant, and since $\cosh x>\sinh x\ge 0$, $\operatorname{sinh}^2 x-\operatorname{cosh}^2 x$ can’t be $1$. That leaves only the right choice, $\operatorname{cosh}^2 x-\operatorname{sinh}^2 x=1$.
If you’re not visually oriented, my advice is simply to learn the exponential formulas and think of the identity in the way suggested by anon in his answer.
A: Use:
$$\cos(ix)^2 + \sin(ix)^2 = 1$$
$$\cosh(x)^2 + i^2 \sinh(x)^2 = 1$$
$$\cosh(x)^2 + i^2 \sinh(x)^2 = 1$$
$$\cosh(x)^2 -\sinh(x)^2 = 1$$
