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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

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4 Answers 4

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

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.

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

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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.

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    $\begingroup$ 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
    Commented Apr 13, 2018 at 11:48
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    $\begingroup$ @Botond thanks for the edit though :) $\endgroup$
    – Mr Pie
    Commented Apr 13, 2018 at 11:49
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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 $\cos^2x+\sin^2x=1$.

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    $\begingroup$ 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
    Commented Apr 16, 2018 at 0:47
  • $\begingroup$ @user477343 Thanks for your correction! $\endgroup$
    – TheSimpliFire
    Commented Apr 16, 2018 at 6:49
  • $\begingroup$ No problem :) ${}$ $\endgroup$
    – Mr Pie
    Commented Apr 16, 2018 at 7:53

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