How do you prove \cosh^2x-\sinh^2x=1?

Substitute \frac12(e^x\pm e^{-x}), expand, and the e^{\pm2x} terms cancel leaving \frac14(4).

How does it differ from \cos^2\theta+\sin^2\theta=1?

It has a minus sign: \cosh^2x-\sinh^2x=1.

Which sign of root is used for \cosh x?

The positive root, as \cosh x\ge1.

\tanh^2x in terms of \sinh x only?

\tanh^2x=\frac{\sinh^2x}{1+\sinh^2x}

If \sinh x=\frac34, find \cosh x and \tanh x.

\cosh x=\frac54 and \tanh x=\frac35

Basic hyperbolic identitiesAQA A-Level Further Maths: Flashcards

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$\tanh x$ in terms of $\sinh x$ and $\cosh x$?

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$\tanh x$ in terms of $\sinh x$ and $\cosh x$ ?: $\tanh x=\frac{\sinh x}{\cosh x}$
The basic hyperbolic identity?: $\cosh^2x-\sinh^2x=1$
How do you prove $\cosh^2x-\sinh^2x=1$ ?: Substitute $\frac12(e^x\pm e^{-x})$ , expand, and the $e^{\pm2x}$ terms cancel leaving $\frac14(4)$ .
How does it differ from $\cos^2\theta+\sin^2\theta=1$ ?: It has a minus sign: $\cosh^2x-\sinh^2x=1$ .
$\cosh^2x$ in terms of $\sinh x$ ?: $\cosh^2x=1+\sinh^2x$
$\sinh^2x$ in terms of $\cosh x$ ?: $\sinh^2x=\cosh^2x-1$
Which sign of root is used for $\cosh x$ ?: The positive root, as $\cosh x\ge1$ .
Result of dividing the identity by $\cosh^2x$ ?: $1-\tanh^2x=\frac{1}{\cosh^2x}$
$\tanh^2x$ in terms of $\sinh x$ only?: $\tanh^2x=\frac{\sinh^2x}{1+\sinh^2x}$
Value of $\cosh x+\sinh x$ ?: $e^x$
If $\cosh x+\sinh x=k$ , what is $\cosh x-\sinh x$ ?: $\frac1k$
If $\sinh x=\frac34$ , find $\cosh x$ and $\tanh x$ .: $\cosh x=\frac54$ and $\tanh x=\frac35$

Exam questions on Basic hyperbolic identities

A real number $x$ satisfies $\sinh x=\frac34$ .
Find the exact value of $\cosh^2x+\sinh^2x$ .2 marks
A real number $x>0$ satisfies $\tanh x=\frac45$ .
Hence find the exact value of $x$ .2 marks
Hyperbolic functions are defined by $\cosh x=\frac12(e^x+e^{-x})$ and $\sinh x=\frac12(e^x-e^{-x})$ .
Prove that $\cosh^2x-\sinh^2x=1$ for all real $x$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).