Basic hyperbolic identitiesAQA A-Level Further Maths: Revision notes

The hyperbolic tangent is defined by $\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}.$ This mirrors $\tan\theta=\frac{\sin\theta}{\cos\theta}$. Because $\cosh x\ge1$ is never zero, $\tanh x$ exists for every real $x$. So if you know $\sinh x$ and $\cosh x$ you find $\tanh x$ by division; if you know $\tanh x$ and $\cosh x$ then $\sinh x=\tanh x\cosh x$. For example, with $\sinh x=\frac34$ and $\cosh x=\frac54$, $\tanh x=\frac35$.

Using the exponential definitions, $\cosh^2x-\sinh^2x=\frac14(e^x+e^{-x})^2-\frac14(e^x-e^{-x})^2=\frac14\left(e^{2x}+2+e^{-2x}\right)-\frac14\left(e^{2x}-2+e^{-2x}\right)=\frac14(4)=1.$ This is the hyperbolic counterpart of $\cos^2\theta+\sin^2\theta=1$, but with a minus sign. It is why the points $(\cosh t,\sinh t)$ lie on the hyperbola $x^2-y^2=1$, giving the functions their name. To prove it in an exam, substitute the exponentials, expand carefully and show that the $e^{\pm2x}$ terms cancel.

Rearranging the identity gives $\cosh^2x=1+\sinh^2x$ and $\sinh^2x=\cosh^2x-1$. Because $\cosh x\ge1>0$, always take the positive root for $\cosh x$; the sign of $\sinh x$ matches the sign of $x$. Example: $\sinh x=\frac34$. Then $\cosh^2x=1+\frac{9}{16}=\frac{25}{16}$, so $\cosh x=\frac54$ and $\tanh x=\frac{3}{5}$. Example: show $\tanh^2x=\frac{\sinh^2x}{1+\sinh^2x}$ by writing $\tanh^2x=\frac{\sinh^2x}{\cosh^2x}$ and replacing $\cosh^2x$ with $1+\sinh^2x$.

Divide $\cosh^2x-\sinh^2x=1$ through by $\cosh^2x$: $1-\tanh^2x=\frac{1}{\cosh^2x}.$ This lets you start from $\tanh x$. Example: $\tanh x=\frac45$ with $x>0$. Then $\frac{1}{\cosh^2x}=1-\frac{16}{25}=\frac{9}{25}$, so $\cosh^2x=\frac{25}{9}$ and $\cosh x=\frac53$. Then $\sinh x=\tanh x\cosh x=\frac43$.

Since $\cosh^2x-\sinh^2x$ is a difference of two squares, $(\cosh x+\sinh x)(\cosh x-\sinh x)=1$. Also $\cosh x+\sinh x=e^x$ and $\cosh x-\sinh x=e^{-x}$, whose product is $1$ as expected. If you are told $\cosh x+\sinh x=4$, then $\cosh x-\sinh x=\frac14$. Adding the two gives $2\cosh x=\frac{17}{4}$, so $\cosh x=\frac{17}{8}$; subtracting gives $\sinh x=\frac{15}{8}$. You can also find $x=\ln4$.