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

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    $\begingroup$ Show your work @kayla $\endgroup$
    – nmasanta
    Aug 5, 2020 at 2:50
  • $\begingroup$ Some useful identities are here: en.wikipedia.org/wiki/Hyperbolic_functions $\endgroup$ Aug 5, 2020 at 2:53
  • $\begingroup$ With $\sinh x = 8/14 = 4/7,$ then yes, $\cosh x = \frac{\sqrt{65}}{7}.$ $\endgroup$
    – mjw
    Aug 5, 2020 at 3:44
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    $\begingroup$ Please do not delete the context of your question. Context is essential for understanding your question and the answers that depend on it. $\endgroup$
    – Toby Mak
    Aug 5, 2020 at 8:37
  • $\begingroup$ @Kayla Do you want to ask a new question for the parts $\cos 2x$ and $\tan 2x$? This part of the question has nothing to do with hyperbolic functions and thus would be better off as its own question? As always, please show the same working out as you have shown here. $\endgroup$
    – Toby Mak
    Aug 10, 2020 at 12:09

3 Answers 3

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

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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 $$\cosh x = \frac{\sqrt{65}}{7}.$$

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  • $\begingroup$ Yeah sinh^2 x = 8/14 sorry $\endgroup$
    – user813939
    Aug 5, 2020 at 3:30
  • $\begingroup$ Okay, so we'll stay with $\sinh x = \frac{8}{14} = \frac{4}{7}.$ That's what you mean, yes. $\endgroup$
    – mjw
    Aug 5, 2020 at 3:37
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$$\sinh x =\frac{e^x-e^{-x}}{2}= \frac{4}{7}$$ 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.)

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