# 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?

-

One approach is to remember that $\cosh x = \cos(ix)$ and $\sinh x = -i \sin (ix)$, which reduces this to $\sin^2 z + \cos^2 z = 1$.

-
Bonus: For the purposes of this identity, you don't even have to remember whether or not there's a minus sign in $\sinh x = -i \sin(ix)$. –  Nate Eldredge Feb 10 '12 at 3:50

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 $y$.

-
Nitpick: The functions $\cosh$ and $\sinh$ are to the right half of the hyperbola $x^2 - y^2 = 1$... –  Rahul Jan 28 '12 at 21:26

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.

-

remember $(a+b)^2-(a-b)^2=4ab$ and $e^xe^{-x}=1$

-

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.

-

Use:

$$\cos(ix)^2 + \sin(ix)^2 = 1$$

$$\cosh(x)^2 + i^2 \sinh(x)^2 = 1$$

$$\cosh(x)^2 + i^2 \sinh(x)^2 = 1$$

$$\cosh(x)^2 -\sinh(x)^2 = 1$$

-