7
$\begingroup$

Does Eulers formula give $$e^{-ix}=\cos(x) -i\sin(x)$$

I know that $$e^{ix}=\cos(x)+i\sin(x)$$ But how does it work when we have a $-$ in front

$\endgroup$
15
$\begingroup$

$$e^{ix}=\cos(x)+i\sin(x)\tag{1}$$ $$e^{-ix}=\cos(x)-i\sin(x)\tag{2}$$

To get from $(1)$ to $(2)$, you just replace $x$ with $-x$. You get the $-$ in front because $\sin(-x)=-\sin(x)$, but the identity is still the same.

$\endgroup$
13
$\begingroup$

The $\sin$ function is odd, so $\sin(-x)=-\sin(x)$.
The $\cos$ function is even, so $\cos(-x)=\cos(x)$.
Using these and Euler's formula, we can get that $$e^{-ix}=e^{i(-x)}=i\sin(-x)+\cos(-x)=-i\sin(x)+\cos(x)$$ If you are not comfortable with it:
Let $u=-x$, then $$e^{-ix}=e^{iu}=i\sin(u)+\cos(u)=i\sin(-x)+\cos(-x)=-i\sin(x)+\cos(x)$$ And you can express the $\sin$ and $\cos$ in the following way: $$e^{ix}+e^{-ix}=(i\sin(x)+\cos(x))+(-i\sin(x)+\cos(x))$$ $$e^{ix}+e^{-ix}=i\sin(x)+\cos(x)+-i\sin(x)+\cos(x)$$ $$e^{ix}+e^{-ix}=2\cos(x)$$ $$\frac{e^{ix}+e^{-ix}}{2}=\cos(x)$$ And $$e^{ix}-e^{-ix}=(i\sin(x)+\cos(x))-(-i\sin(x)+\cos(x))$$ $$e^{ix}-e^{-ix}=i\sin(x)+\cos(x)+i\sin(x)-\cos(x)$$ $$e^{ix}-e^{-ix}=2i\sin(x)$$ $$\frac{e^{ix}-e^{-ix}}{2i}=\sin(x)$$

$\endgroup$
5
$\begingroup$

Remember that $$\begin{align}-i &=-\sqrt{-1} \\ &=(-1)\sqrt{-1} \\ &=i^2\cdot i \\ &= i^3\end{align}$$ Thus, if

$e^{ix} = \cos(x) + i\sin(x),$ then $e^{-ix} = (e^{ix})^{-1}.$

Therefore,

$$\boxed{ \ e^{-ix} = \frac{1}{\cos(x) + i\sin(x)}. \ }$$


Edit:

As @Botond suggested, by using a conjugate method (used to rationalise denominators) as mentioned in his comment below, we get the much nicer result, $$e^{-ix} = \frac{\cos(x) - i\sin(x)}{\cos(x)^2 + \sin(x)^2}.$$ Since $\cos(x)^2 + \sin(x)^2 = 1$ then yes, correct indeed.

$\endgroup$
  • 1
    $\begingroup$ I think you could multiply your result by $1=\frac{\cos(x)-i\sin(x)}{\cos(x)-i\sin(x)}$ to get a bit nicer result. $\endgroup$ – Botond Apr 13 '18 at 11:48
  • 1
    $\begingroup$ @Botond thanks for the edit though :) $\endgroup$ – Mr Pie Apr 13 '18 at 11:49
3
$\begingroup$

From the property that for any $z_1,z_2\in\mathbb{C}$, $$e^{z_1}\cdot e^{z_2}=e^{z_1+z_2}$$ we have that $$e^{ix}\cdot e^{-ix}=e^{ix-ix}=e^0=1$$ so $$e^{-ix}=\frac1{e^{ix}}=\frac{\overline{e^{ix}}}{e^{ix}\cdot\overline{e^{ix}}}=\frac{\overline{e^{ix}}}{|e^{ix}|^2}=\overline{e^{ix}}$$ using that $\cos^2x+\sin^2x=1$.

$\endgroup$
  • 1
    $\begingroup$ You made a typo in your second equation. It should be $e^{ix-ix}$ instead of $e^{-ix-ix}$. Either way, $(+1)$. $\endgroup$ – Mr Pie Apr 16 '18 at 0:47
  • $\begingroup$ @user477343 Thanks for your correction! $\endgroup$ – TheSimpliFire Apr 16 '18 at 6:49
  • $\begingroup$ No problem :) ${}$ $\endgroup$ – Mr Pie Apr 16 '18 at 7:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.