Further hyperbolic identities and proofsAQA A-Level Further Maths: Revision notes

Hyperbolic functions are defined by $\sinh x=\frac{e^x-e^{-x}}{2}$ and $\cosh x=\frac{e^x+e^{-x}}{2}$, with $\tanh x=\frac{\sinh x}{\cosh x}$, $\operatorname{sech}x=\frac1{\cosh x}$, $\operatorname{cosech}x=\frac1{\sinh x}$ and $\coth x=\frac{\cosh x}{\sinh x}$. Squaring and subtracting gives the fundamental identity $\cosh^2x-\sinh^2x=1.$ Compare with $\cos^2x+\sin^2x=1$: the sign is different, because $(\cosh x,\sinh x)$ lies on a hyperbola, not a circle. Every other identity in this topic is a rearrangement or a division of this one.

Divide $\cosh^2x-\sinh^2x=1$ by $\cosh^2x$: $1-\tanh^2x=\operatorname{sech}^2x.$ Divide it by $\sinh^2x$: $\coth^2x-1=\operatorname{cosech}^2x.$ These are the AQA identities $\operatorname{sech}^2x=1-\tanh^2x$ and $\operatorname{cosech}^2x=\coth^2x-1$. Example: if $\tanh x=\frac35$, then $\operatorname{sech}^2x=1-\frac9{25}=\frac{16}{25}$, so $\cosh x=\frac54$ (positive, since $\cosh x\geq1$) and $\sinh x=\tanh x\cosh x=\frac34$.

Two double-angle results are on the specification: $\sinh 2x=2\sinh x\cosh x,\qquad \cosh 2x=\cosh^2x+\sinh^2x.$ To prove the second, expand the exponentials: $\cosh^2x+\sinh^2x=\frac{(e^x+e^{-x})^2+(e^x-e^{-x})^2}{4}=\frac{2e^{2x}+2e^{-2x}}{4}=\cosh 2x$. Combining it with $\cosh^2x-\sinh^2x=1$ gives two more forms: $\cosh 2x=1+2\sinh^2x=2\cosh^2x-1.$ Dividing gives $\tanh 2x=\frac{2\tanh x}{1+\tanh^2x}$. Unlike $\cos 2x$, there is no minus sign in $\cosh^2x+\sinh^2x$.

To prove an identity, start from the more complicated side and transform it into the other, showing every step. Two reliable methods:

• Use known identities. Replace $\sinh 2x$, $\cosh 2x$ and any $1$ using $\cosh^2x-\sinh^2x=1$, then simplify.
• Use exponentials. Replace each function with its $e^x$ form and expand. This always works when the identities are not obvious. Worked example: prove $\tanh 2x\equiv\frac{2\tanh x}{1+\tanh^2x}$. $\tanh 2x=\frac{2\sinh x\cosh x}{\cosh^2x+\sinh^2x}$. Divide top and bottom by $\cosh^2x$ to get $\frac{2\tanh x}{1+\tanh^2x}$. End with a conclusion. If the question says "show that", the final line must be the printed result.

Identities turn an equation into one in a single function. To solve $\cosh 2x-3\sinh x=1$, use $\cosh 2x=1+2\sinh^2x$: $2\sinh^2x-3\sinh x=0$, so $\sinh x=0$ or $\sinh x=\frac32$. Then $x=0$ or, from $e^x-e^{-x}=3$, $e^{2x}-3e^x-1=0$, so $e^x=\frac{3+\sqrt{13}}{2}$ and $x=\ln\frac{3+\sqrt{13}}2$. Check each root against the range of the function: $\cosh x\geq1$, $-1<\tanh x<1$ and $e^x>0$. If $\tanh 2x=\frac45$ gives $2t^2-5t+2=0$, then $t=\frac12$ is valid but $t=2$ must be rejected. To evaluate at a value such as $x=\frac12\ln5$, use the identity first, then the exponential definitions with $e^{\ln5}=5$.