# Thread: Hyperbolic trig identities?

1. ## Hyperbolic trig identities?

Hi,

I'm not sure how or which identities to use to solve the following question:

Express $sinh^5x$ in terms of hyperbolic sines of multiples of $x$.

The answer is: $\frac{1}{16}(sinh\ x-5sinh\ 3x+10sinh\ x)$

I have a fair few similar problems, so if you could please show me the steps for this one I should hopefully be able to solve the others.

Thanks heaps for your help.

2. ## Re: Hyperbolic trig identities?

Originally Posted by Stroodle
Hi,

I'm not sure how or which identities to use to solve the following question:

Express $sinh^5x$ in terms of hyperbolic sines of multiples of $x$.

The answer is: $\frac{1}{16}(sinh\ x-5sinh\ 3x+10sinh\ x)$

I have a fair few similar problems, so if you could please show me the steps for this one I should hopefully be able to solve the others.

Thanks heaps for your help.

$\sinh{x}=\frac{e^x-e^{-x}}{2}$

$\sinh^5{x}=(\frac{e^x-e^{-x}}{2})^5$

$=\frac{1}{32}(e^{5x}-5e^{3x}+10e^{x}-10e^{-x}+5e^{-3x}-e^{-5x})$

$=\frac{1}{32}((e^{5x}-e^{-5x})-5(e^{3x}-e^{-3x})-10(e^{x}-e^{-x}))$

So, what can you say?

3. ## Re: Hyperbolic trig identities?

Awesome. Thanks for that.
I think I'm meant to use the hyperbolic trig identities though (I might be wrong, but it's in that section of my text).

4. ## Re: Hyperbolic trig identities?

Originally Posted by Stroodle
Awesome. Thanks for that.
I think I'm meant to use the hyperbolic trig identities though (I might be wrong, but it's in that section of my text).

By definition:

$\sinh{x}=\frac{e^x-e^{-x}}{2}$

$\cosh{x}=\frac{e^x+e^{-x}}{2}$

Now, with the above you can prove:

1) $\cosh^2{x}-\sinh^2{x}=1$

2) $\sinh{2x}=2\sinh{x}\cosh{x}$

3) $1-\tanh^2{x}=\frac{1}{\cosh^2{x}}$

4) $\cosh{x+y}=\cosh{x}\cosh{x}+\sinh{x}\sinh{y}$

5) $\sinh{x+y}=\sinh{x+y=\sinh{x}\cosh{y}+\cosh{x}\sin h{y}$

6) $\sinh{x}+\sin{y}=2\sinh{(\frac{x+y}{2})}\cosh (\frac{x-y}{2})$

AND MORE...

Try to prove 1-6, try to find more identities...