1
$\begingroup$

Prove that $\cosh x-1\equiv \frac12(e^{0.5x}-e^{-0.5x})^2$

I'm stuck on what appears to be the last step, please could someone explain where I have made a mistake?

\begin{align} \frac12(e^{0.5x}-e^{-0.5x})^2 & \equiv \frac12(\sinh(0.5x))^2 \\ & \equiv \frac12\sinh^2(0.5x) \\ & \equiv \frac12(\cosh^2(0.5x) -1) \\ & \equiv \frac12((\frac{e^{0.5x}+e^{-0.5x}}{2})^2 -1) \\ & \equiv \frac12((\frac{e^x+e^{-x}+2}4) -1) \\ & \equiv \frac14((\frac{e^x+e^{-x}+2}2)-2) \\ & \equiv \frac14(\frac{e^x+e^{-x}}2+\frac22-2) \\ & \equiv \frac14(\frac{e^x+e^{-x}}{2}-1) \\ & \equiv \frac14(\cosh x-1) \\ \end{align}

which is a quarter of what I am trying to prove?

$\endgroup$
1
  • $\begingroup$ $$e^{\frac12x}-e^{-\frac12x}=\color{red}2\sinh \frac12x$$ $\endgroup$ Commented Mar 13, 2018 at 16:31

2 Answers 2

Reset to default
2
$\begingroup$

Be careful with your multiplicative factors! In the very first step, you should have written $$\frac12(e^{0.5x}-e^{-0.5x})^2=\color{red}2(\sinh0.5x)^2$$ With this correction, the rest of your steps check out and you get the correct answer.

$\endgroup$
0
$\begingroup$

Note that $$\begin{align} \frac12(e^{0.5x}-e^{-0.5x})^2 & \equiv \frac12(2\sinh(0.5x))^2 \\ & \equiv 2\sinh^2(0.5x) \\ & \equiv 2(\cosh^2(0.5x) -1) \\ & \equiv 2((\frac{e^{0.5x}+e^{-0.5x}}{2})^2 -1) \\ & \equiv 2((\frac{e^x+e^{-x}+2}4) -1) \\ & \equiv ((\frac{e^x+e^{-x}+2}2)-2) \\ & \equiv (\frac{e^x+e^{-x}}2+\frac22-2) \\ & \equiv (\frac{e^x+e^{-x}}{2}-1) \\ & \equiv (\cosh x-1) \\ \end{align}$$ Which is what you wanted to prove.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.