2
$\begingroup$

I just did a question where I had $sinh^2(x)$

I know this is simply $(sinh(x))^2$ however couldn't work out where the extra 2 came from when working out.

$sinh(x) = \frac{e^x-e^{-x}}{2}$

so

$sinh^2(x) = (\frac{e^x-e^{-x}}{2})^2$

which I figured

= $\frac{e^xe^x-e^{-x}e^{-x}}{4}$

= $\frac{e^{2x}-e^{-2x}}{4}$

However I am told the answer

= $\frac{e^{2x}-2+e^{-2x}}{4}$

And I don't know where the 2 came from

$\endgroup$
1
  • 2
    $\begingroup$ $(a-b)^2 = a^2 - 2ab + b^2$. Your result $a^2-b^2$ is $(a-b)(a+b)$. $\endgroup$ Nov 1, 2013 at 13:06

2 Answers 2

4
$\begingroup$

You are trying to do $(a-b)^2=a^2-b^2$, which is not correct. The correct expansion is $(a-b)^2=a^2-2ab+b^2$ Inserting $a=e^x, b=e^{-x}$, gives the formula you were given. The $2$ comes because $ab=1$

$\endgroup$
3
$\begingroup$

The $2$ is the OI term in the FOIL rule for squaring. Note that $e^x\cdot e^{-x} = 1$ for any $x$. In fact, you can do more here.

$$\frac{e^{2x}-2+e^{-2x}}{4} = -{1\over 2} + {1\over 2}\cosh(2x).$$

$\endgroup$
2
  • $\begingroup$ Sorry I'm not familiar with the FOIL rule. Say if we were not dealing with e, but rather the same equation with y instead. Is my working then correct meaning this rule only applies when we are dealing with e? $\endgroup$
    – user88720
    Nov 1, 2013 at 13:05
  • $\begingroup$ The FOIL rule goes as follows: $(a + b)(c + d)$ = $ac + ad + bc + bd$. $ac$ is the product of the first terms (F), $ad$ is the product of the outer terms (O), $bc$ is the product of the inner terms (I) and $bd$ is the product of the last terms (L). When using this to square the O and I terms are like and are combined, hence OI term. $\endgroup$ Nov 1, 2013 at 13:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .