If $t=\tanh\frac{x}{2}$, prove that $\sinh x = \frac{2t}{1-t^2}$ and $\cosh x = \frac{1+t^2}{1-t^2}$. Hence solve the equation $7\sinh x + 20 \cosh x = 24$.

I have tried starting by writing out $\tanh\frac{x}{2}$ in exponential form and then squaring it but I can't make any progress from this.

  • 1
    $\begingroup$ Do you know any relationships between $\tanh x$ and $\text{sech} x$ for example. $\endgroup$ – Chinny84 Aug 14 '15 at 11:52
  • $\begingroup$ Multiply with $\dfrac{\cosh^2 \frac{x}{2}}{\cosh^2 \frac{x}{2}}$. $\endgroup$ – Daniel Fischer Aug 14 '15 at 11:58

$$\sinh(x)=2\sinh(\frac x 2)\cosh(\frac x2)=\dfrac 2{\dfrac{\cosh^2(\frac x2)-\sinh^2(\frac x2)}{\sinh(\frac x 2)\cosh(\frac x2)}}=\frac 2{\dfrac1 {\tanh(x/2)}-\tanh(x/2)}=\frac{2t}{1-t^2}$$ $$\tanh(x)=\frac{2\tanh(x/2)}{1+\tanh^2(x/2)}=\frac{2t}{1+t^2}$$ $$\cosh(x)=\frac{\sinh(x)}{\tanh(x)}=\frac{1+t^2}{1-t^2}$$


$$7\sinh(x)+20\cosh(x)=24$$ $$44t^2+14t-4=0$$


Note that

(i) tanh($x$) = sinh(x) / cosh($x$)

(ii) 1 - tanh$^2(x$) = 1/ cosh$^2(x$)

(iii) sinh($x + y$) = sinh($x$)cosh($y$) + sinh($y$)cosh($x$)

and apply these to the term

$\frac{2t}{1 - t^2}$ with $t =$ tan($x/2$) to directly obtain your result for sinh($x$). The formula for cosh$(x)$ follows then directly by applying (i) again.


Let $t$=$tanh$$\frac{x}{2}$.

Using Identity $sech^2$$\frac{x}{2}$=1-$t^2$


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

$cosh$ $x$ =2 ($\frac{1}{1-t^2}$)-1

$cosh$ $x$=$\frac{1+t^2}{1-t^2}$ is obtained by simplifying the above.

Using another identity $cosh^2$$\frac{x}{2}$-$sinh^2$$\frac{x}{2}$=1



$sinh $$\frac{x}{2}$=$\frac{t}{\sqrt(1-t^2)}$

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

$sinh$ $x$=$\frac{2t}{1-t^2}$ is obtained by simplifying the above.

Moving to the 2nd part after replacing with $t$ in the equation, you will get

12$t^2$+17$t$-2=0.From here I guess you should solve for $t$. You will get something in terms of $\sqrt385$. What you can do is replace it in $cosh$ $x$ and simplify by $x$ = $\underline+$$ln$($x+\sqrt{1+x^2}$). I have not tried it but still I hope it helps.


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