How to "see" this algebraic transformation used to derive the exponential form of the hyperbolic cosine?

Question

In this page, the exponential form of cosh is derived from the geometric definition (which is based on the unit hyperbola). Here is one step of the derivation, fully expanded out for novices like me:

$$a = ln(b + \sqrt{b^2 - 1})$$

$$e^a = b + \sqrt{b^2 - 1}$$

$${1 \over e^a} = {1 \over b + \sqrt{b^2 - 1}}$$

$$e^a + {1 \over e^a} = b + \sqrt{b^2 - 1} + {1 \over b + \sqrt{b^2 - 1}}$$

$$e^a + {1 \over e^a} = {b(b + \sqrt{b^2 - 1}) + \sqrt{b^2 - 1}(b + \sqrt{b^2 - 1}) + 1 \over b + \sqrt{b^2 - 1}}$$

$$e^a + {1 \over e^a} = {b^2 + b\sqrt{b^2 - 1} + b\sqrt{b^2 - 1} + b^2 - 1 + 1 \over b + \sqrt{b^2 - 1}}$$

$$e^a + {1 \over e^a} = {2b^2 + 2b\sqrt{b^2 - 1} \over b + \sqrt{b^2 - 1}}$$

$$e^a + {1 \over e^a} = 2b$$

This is all well and good, but for the life of me I can't see: how would someone ever know in advance that adding $b + \sqrt{b^2 - 1}$ to its own reciprocal would clean things up so nicely? Is this just a "special case" which needs to be memorized, or is there some general principle behind it which can be used when solving similar equations?

Answer 1

Another "interesting" way to solve this equation, and which may be more easily explainable how to "see", might be to intuit that the right-hand side, with its sum of a number and a square root with a square of that number underneath it kind of resembles a quadratic formula:

$$b + \sqrt{b^2 - 1}$$

has kinda the form of

$$\frac{-B + \sqrt{B^2 - 4AC}}{2A}$$

Namely, note the

$$-B + \sqrt{B^2 - \cdots}$$

in particular, exactly rewritable with the right choices of $A$, $B$, and $C$ as

$$\frac{-(-b) + \sqrt{(-b)^2 - 4\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)}}{2\left(\frac{1}{2}\right)}$$

(think: "How do I cancel the $4$ down to a $1$?" "Ooh, I can also end up removing the $2$ for 'free' by noting $4 = 2 \cdot 2$!" "Now what about the wee problem of $B$ being negative on the outside which doesn't quiiiiiite match?") which is the quadratic formula for the quadratic equation

$$\frac{1}{2} x^2 + (-b) x + \frac{1}{2} = 0$$

and then, since that quadratic formula was just the original right-hand side rewritten, and the quadratic formula gives "$x$", and the original equals $e^a$, we have $x = e^a$, which gives

$$\frac{1}{2} (e^a)^2 + (-b)(e^a) + \frac{1}{2} = 0$$

divide out $e^a$,

$$\frac{1}{2} (e^a) + (-b) + \frac{1}{2} (e^{-a}) = 0$$

and you can pull out $b$ yourself to get

$$b = \frac{e^a + e^{-a}}{2} = \cosh\ a$$.

Insofar as the original argument from the website is concerned, I suspect what the author did was that they already knew the form that $\cosh$ is supposed to take (who writing that wouldn't?) and just used that as a guide. Once you have $e^a$, you know you need to add $e^{-a}$ (i.e. $\frac{1}{e^a}$). The method above, though, is what you might do if you hypothetically never knew what $\cosh$ was supposed to look like and were deriving it afresh.

Answer 2

Let $$\cosh(a)=b$$ and solve for $a$

$$e^a+e^{-a}=2b\implies e^{2a}-2be^a+1=0$$

Solve for $e^a$ with quadratic formula $$ e^a= b\pm \sqrt {b^2-1}$$

Choose the plus sign and take logarithm and you have the answer.

Answer 3

Lets call $c=\sqrt{(b^2-1)}$. Note that $b^2-c^2=1$ so $(b+c)(b-c)=1$ and $b-c={1\over b+c}$ , so that $b+c +{1\over b+c} = b+c + b-c =2b$