Given $\sinh x$, find the exact value of $\cosh x, \cos 2x$ and $\tan 2x$

Question

Given $\sinh x = 8/14$, find the exact value of $\cosh x, \cos 2x$ and $\tan 2x$.

I have been getting two answers which has made me confused. I keep getting $\sqrt{65/7}$ or $\sqrt{4/7}.$

That's what I got: $$\cosh^2 x - \sinh^2 x = 1$$ $$\cosh^2 x = 1 + \sinh^2 x$$ $$\cosh^2 x = 1 +\left(\frac{8}{14}\right)^2$$ $$\cosh^2 x=\frac{65}{49}$$ $$\cosh (x) = \sqrt{\frac{65}{49}}$$ $$\cosh x = \frac{\sqrt{65}}{7}$$

Answer 1

$$\sinh x =\frac{e^x-e^{-x}}{2}= \frac{4}{7}$$ We want $$\cosh x = \frac{e^x+e^{-x}}{2} = q$$ Adding these: $$e^x = \frac{4}{7}+q$$ Subtracting the first equation from the second: $$e^{-x} = q - \frac{4}{7}$$ So $$\frac{4}{7}+q = \frac{1}{q-\frac{4}{7}}$$ $$q^2 - \left(\frac{4}{7}\right)^2 =1$$ $$q^2 =1+\frac{16}{49}$$ $$\cosh x = q=\frac{\sqrt{65}}{7}$$(q must be positive because each exponential is positive.)

Answer 2

You are right! Just use $$\cosh^2 x = 1+\sinh^2 x.$$

Now it only amounts to keeping track of what $\sinh x$ is and carefully doing the calculation. Is it $\sinh^2 x = \left(\frac{8}{13}\right)^2$ or is it $\sinh^2 x = \left(\frac{8}{14}\right)^2$?

UPDATE: With $\sinh x = \frac{4}{7}$:

$$\cosh^2 x = 1+ \left(\frac{4}{7}\right)^2 = 1+ \frac{16}{49}$$ so $$\cosh x = \frac{\sqrt{65}}{7}.$$

Answer 3

$$\cosh^2 x = 1 + \sinh^2 x.$$ error was here $$\rightarrow \cosh^2 x = 1 +(8/13)^2.$$ $$\cosh^2 x=(233/169).$$ $$\cosh (x) = \sqrt {233}/13.$$