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.
Best Answer
$$\sinh(x)=2\sinh(\frac x 2)\cosh(\frac x2)=\dfrac 2{\dfrac{\cosh^2(\frac x2)-\sinh^2(\frac x2)}{\sinh(\frac x 2)\cosh(\frac x2)}}=\frac 2{\dfrac1 {\tanh(x/2)}-\tanh(x/2)}=\frac{2t}{1-t^2}$$ $$\tanh(x)=\frac{2\tanh(x/2)}{1+\tanh^2(x/2)}=\frac{2t}{1+t^2}$$ $$\cosh(x)=\frac{\sinh(x)}{\tanh(x)}=\frac{1+t^2}{1-t^2}$$
Hence:
$$7\sinh(x)+20\cosh(x)=24$$ $$44t^2+14t-4=0$$