2022-01-17

Why is ${e}^{i\theta }=\mathrm{cos}\left(\theta \right)+i\mathrm{sin}\left(\theta \right)?$

nick1337

Expert

Step 1
Consider the function
$f\left(x\right)=\frac{\mathrm{cos}x+i\mathrm{sin}x}{{e}^{ix}}$
which converts a real x to an intricate number. Utilizing the quotient rule, determine the derivative with respect to x:
${f}^{\prime }\left(x\right)=\frac{\left(-\mathrm{sin}x+i×\mathrm{cos}x\right)×{e}^{ix}-\left(\mathrm{cos}x+i×\in x\right)×i×{e}^{ix}}{{e}^{i{x}^{2}}}$
Reordering the numerator proves: $f\prime \left(x\right)=0$, in other words: f is a constant function; all reals are mapped to the same complex number. Which number? OK, try $x=0$; we know: $\mathrm{cos}0=1,\mathrm{sin}0=0,{e}^{0}=1;$ so $f\left(0\right)=1$, so $f\left(x\right)=1$ for all x, or
$1=\frac{\mathrm{cos}x+i×\mathrm{sin}x}{{e}^{ix}}$
or
${e}^{ix}=\mathrm{cos}x+i×\mathrm{sin}x$
QED.

star233

Expert

Step 1
The reason this is true for most students nowadays is more so because their teachers told them this is true than anything else. But if you were asking to show that it is true using mathematics, then that is also possible to do.
We start by letting
$z=\mathrm{cos}\theta +i\mathrm{sin}\theta$
$⇒{z}^{\prime }=-\mathrm{sin}\theta +i\mathrm{cos}\theta$
Recall that ${i}^{2}=\left(-1,0\right)$. That means we can replace the minus one with ${i}^{2}$
${z}^{\prime }={i}^{2}\mathrm{sin}\theta +i\mathrm{cos}\theta$
Factoring out an i , we are left with:
${z}^{\prime }=i\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)$
Recall that
$\mathrm{cos}\theta +i\mathrm{sin}\theta =z\phantom{\rule{0ex}{0ex}}{z}^{\prime }=iz$
How interesting. We see that this function must be such that its derivative is equal to itself multiplied by some constant. Doesn’t that sound oddly similar to the exponential function? Let’s keep going.
$\frac{{z}^{\prime }}{z}=i$
We can now integrate both sides because we want to remove all of our derivatives.
$\int \frac{{z}^{\prime }}{z}d\theta =\int id\theta$
The integral on the left is the world famous natural logarithm. The integral on the right is trivial because we are just integrating some constant function.
$\mathrm{ln}z=i\theta +C$
Recall now the definition of the logarithm.
$z={e}^{i\theta +C}$
Wow, looks like we’re starting to get there!
$z={e}^{i\theta }{e}^{C}$
I rewrote ${e}^{C}$ as just C because they’re both just some constants of integration.
$⇒z=C×{e}^{i\theta }$
We’re so close, but we have to find the value for the constant of integration before we go anywhere else. We’ll start by replacing z for what it is substituting.
$\mathrm{cos}\theta +i\mathrm{sin}\theta =C×{e}^{i\theta }$
Let's let $\theta =0$. In doing so, we see that a lot of things cancel nicely.
$\mathrm{cos}0+i\mathrm{sin}0=C×{e}^{0}$
$1=C$
Now we know the value for the constant, we can go back to the equation we had earlier.
$z=C×{e}^{i\theta }$
Now, to make all of our proper substitutions.
$\mathrm{cos}\theta +i\mathrm{sin}\theta ={e}^{i\theta }$
And now we know why this is true.

alenahelenash

Expert

Step 1 The most common way is to calculate the Taylor series for each: $\mathrm{exp}\left(z\right)=1+\frac{{z}^{1}}{1!}+\frac{{z}^{2}}{2!}+\frac{{z}^{3}}{3!}+\frac{{z}^{4}}{4!}+\frac{{z}^{5}}{5!}+\cdots$ $\mathrm{cos}z=1-\frac{{z}^{2}}{2!}+\frac{{z}^{4}}{4!}-\frac{{z}^{6}}{6!}+\frac{{z}^{8}}{8!}-\frac{{z}^{10}}{10!}+\cdots$ $\mathrm{sin}z=\frac{{z}^{1}}{1!}=\frac{{z}^{3}}{3!}+\frac{{z}^{5}}{5!}-\frac{{z}^{7}}{7!}+\frac{{z}^{9}}{9!}-\frac{{z}^{11}}{11!}+\cdots$ Now calculate the series for $\mathrm{exp}\left(iz\right)$ and see how it compares when you add the series for $\mathrm{cos}z$ and $i\mathrm{sin}z$ .