# Can hyperbolic sine and cosine be combined into a single function of shifted argument?

For trigonometric functions we have a nice identity:

$$A\cos x+B\sin x=\sqrt{A^2+B^2}\sin(x+\operatorname{atan2}(A,B)).\tag1$$

At the core of it is the well-known identity of

$$\sin^2x+\cos^2y=1,\tag2$$

which allows us to view $\frac A{\sqrt{A^2+B^2}}$ as sine and $\frac B{\sqrt{A^2+B^2}}$ as cosine of some angle (same angle for both expressions).

The analogous identity for hyperbolic functions is

$$\cosh^2x-\sinh^2x=1.\tag3$$

But I can't seem to be able to use it to find hyperbolic analogue for $(1)$. The difference of squares seems to be not constraining enough to allow for it, unlike the sum.

So is there any identity for hyperbolic functions, that would be analogous to $(1)$?

• You could've hit it head on with the definitions of $\cosh$ and $\sinh$. – Simply Beautiful Art Feb 13 '16 at 20:29

## 2 Answers

For example,

Given $$A\cosh x+B\sinh x=R \cosh (x+\alpha)=R\cosh x\cosh \alpha+R\sinh x\sinh\alpha$$

Then $$A=R\cosh \alpha,B=R\sinh \alpha$$

So $$\tanh\alpha=\frac BA$$ and$$R^2=A^2-B^2$$

This is assuming $A\geq B$. The other compound angle identities can be used in a similar way.

For reference, the complete identity (which appears to be not as nice as $(1)$ in the OP) is

$$A\cosh x+B\sinh x=\begin{cases} A\sqrt{1-\frac{B^2}{A^2}}\cosh\left(x+\operatorname{artanh}\frac BA\right)& \text{if }A^2>B^2\\ B\sqrt{1-\frac{A^2}{B^2}}\sinh\left(x+\operatorname{artanh}\frac AB\right)& \text{if }A^2<B^2\\ A\exp(x)& \text{if }A=B \end{cases}.$$

Here $A$ in the first case and $B$ in the second case are taken out of radicals to take their sign into account.