Given $\sinh x$, find the exact value of $\cosh x, \cos 2x$ and $\tan 2x$ 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:  $$\cosh^2 x - \sinh^2 x = 1$$ $$\cosh^2 x = 1 + \sinh^2 x$$ $$\cosh^2 x = 1 +\left(\frac{8}{14}\right)^2$$ $$\cosh^2 x=\frac{65}{49}$$ $$\cosh (x) = \sqrt{\frac{65}{49}}$$ $$\cosh x = \frac{\sqrt{65}}{7}$$

$$\sinh x =\frac{e^x-e^{-x}}{2}= \frac{4}{7}$$
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.)
