# How is $2 + e^{t^2} - e^{-t^2}$ Equal to $(e^t + e^{-t})^2$?

Could someone please break down explain how the following is true. I'm not sure what happens to the 2 during simplification:

$2 + e^{t^2} - e^{-t^2} = \left(e^t + e^{-t}\right)^2$

Sorry for such a general question, but having a good understanding of this would be very helpful.

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@AustinMohr Yep, I made a silly typo. – Oliver Spryn Oct 11 '12 at 0:35
still, putting $t=0$ gives $2$ on the LHS, and $4$ on the RHS... There is no way an identity like this is possible, $\exp(t^2)$ grows far too quickly compared to $\exp(t)$. – Olivier Bégassat Oct 11 '12 at 0:39
As the answers indicate, you likely made two transcription errors: using $t^2$ rather than $2t$ as the exponents and using $-$ instead of $+$ on the left of your equation. – Rick Decker Oct 11 '12 at 0:42
Probably you should write parentheses ... $(e^t)^2$ since without them the convention is not that. – GEdgar Oct 11 '12 at 1:46

## 2 Answers

As with most factoring questions, it's much easier to work backwards. $$\left(e^t + e^{-t}\right)^2 = (e^t)^2 + 2(e^t)(e^{-t}) + (e^{-t})^2$$ So you can see that the $2$ comes from the fact that $e^t$ and $e^{-t}$ are reciprocals. This occurs more often than you would imagine, so it might be useful to take note of this property.

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Oh... Thank you for making that clear. The \frac{e^t}{e^t} cancel to give one. That is clear in the way you broke it up. Unfortunately, I was given the longer expression to start with, and had to see if it simplified, so I didn't have the opportunity to work backwards. – Oliver Spryn Oct 11 '12 at 0:42

The 2 is the OI term in undoing the FOIL rule. Look at the reverse $$(e^t + e^{-t})^2 = e^{2t} + 2e^te^{-t} + e^{-2t} = 2 + e^{2t} + e^{-2t}.$$

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