Hyperbolic functions and identitiesEdexcel International A Level Further Maths: Revision notes

The hyperbolic functions are defined from $e^x$: $\sinh x=\frac{e^x-e^{-x}}{2},\quad\cosh x=\frac{e^x+e^{-x}}{2},\quad\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}.$ The reciprocals are $\operatorname{sech}x=\frac{1}{\cosh x}=\frac{2}{e^x+e^{-x}}$, $\operatorname{cosech}x=\frac{1}{\sinh x}=\frac{2}{e^x-e^{-x}}$ and $\coth x=\frac{1}{\tanh x}=\frac{\cosh x}{\sinh x}$. Multiplying the top and bottom of $\tanh x$ by $e^x$ gives $\tanh x=\frac{e^{2x}-1}{e^{2x}+1}$. Example: $\cosh(\ln3)=\frac12\left(3+\frac13\right)=\frac53$ and $\sinh(\ln3)=\frac43$, so $\tanh(\ln3)=\frac45$.

• $y=\cosh x$: defined for all real $x$, even ($\cosh(-x)=\cosh x$), range $y\ge1$, minimum value $1$ at $x=0$, U-shaped.
• $y=\sinh x$: odd ($\sinh(-x)=-\sinh x$), range all real numbers, increasing, passes through the origin.
• $y=\tanh x$: odd, range $-1<y<1$, increasing, with asymptotes $y=1$ and $y=-1$.
• $y=\operatorname{sech}x$: even, range $0<y\le1$, maximum $1$ at $x=0$, asymptote $y=0$.
• $y=\operatorname{cosech}x$ and $y=\coth x$: not defined at $x=0$ (vertical asymptote $x=0$); both odd. $\operatorname{cosech}x$ has the $x$-axis as an asymptote and $\coth x$ has $y=\pm1$ as asymptotes, with range $|y|>1$. For large positive $x$, $\sinh x\approx\cosh x\approx\frac12e^x$.

Squaring the definitions and using $e^x\times e^{-x}=1$: $\cosh^2x=\frac{e^{2x}+2+e^{-2x}}{4},\qquad\sinh^2x=\frac{e^{2x}-2+e^{-2x}}{4}.$ Subtracting and adding gives the key identities: $\cosh^2x-\sinh^2x=1,\qquad\cosh^2x+\sinh^2x=\cosh2x.$ Combining them gives $\cosh2x=1+2\sinh^2x=2\cosh^2x-1$. Dividing the first by $\cosh^2x$ gives $1-\tanh^2x=\operatorname{sech}^2x$. These look like the trigonometric identities, but with a sign change: $\cosh^2x-\sinh^2x=1$, not $+$.

Use an identity to turn an equation into a quadratic in a single hyperbolic function. For $\cosh2x+3\sinh x=3$, replace $\cosh2x$ by $1+2\sinh^2x$: $2\sinh^2x+3\sinh x-2=0\ \Rightarrow\ (2\sinh x-1)(\sinh x+2)=0.$ So $\sinh x=\frac12$ or $-2$, giving $x=\sinh^{-1}\left(\frac12\right)=0.481$ or $x=\sinh^{-1}(-2)=-1.44$. Any $\sinh x$ value is allowed, but $\cosh x$ must be at least $1$, so reject a root such as $\cosh x=\frac12$.

Replace $\cosh x$ and $\sinh x$ with their exponential definitions, collect the terms in $e^x$ and $e^{-x}$, and multiply by $e^x$ to get a quadratic in $e^x$. Example: $5\cosh x+3\sinh x=5$ becomes $4e^x+e^{-x}=5$, so $4e^{2x}-5e^x+1=0$ and $(4e^x-1)(e^x-1)=0$. Hence $e^x=\frac14$ or $1$, giving $x=-\ln4$ or $x=0$. Because $e^x>0$, reject any root where $e^x\le0$. If $e^x$ is positive, take natural logs; a negative or zero value means no solution from that root.