You are watching: E^ix+e^-ix

By Euler"s formula,

$e^ix=\cos x+i \sin x$

$e^-ix=\cos x-i\sin x$

then,

$e^ix+e^-ix=2\cos x \Rightarrow \cos x=\frac12(e^ix+e^-ix)$.

Euler"s formula claims that$$e^ix=\cos x+i\sin x$$By instead of $x$ v $-x$ and using the parities that the trigonometric features (the sine is odd and also the cosine even) us get$$e^-ix=\cos x-i\sin x$$Adding these two equations together we get$$e^ix+e^-ix=2\cos x$$and finally$$\frace^ix+e^-ix2=\cos x.$$

$e^ix=\cos x+i \sin x$, for this reason $e^-ix=\cos(-x)+i\sin(-x)=\cos x-i\sin x$. Adding these 2 equations you gain the forced formula.

Let $f(x)=\frac 12 (e^ix+e^-ix)$.

Then that is simple to present that $y=f(x)$ is a solution to $$y""+y=0 \,;\, y(0)=1, y"(0)=0 $$

But this equation has unique solution, and $\cos(x)$ is additionally a systems to this.

Taylor series of $\cos(x)$:$$\beginalign\cos(x)&=\colorgreen1-\fracx^22!+\fracx44!-\cdots\endalign$$

Taylor collection of $e^x$:$$e^x=1+x+\fracx^22!+\fracx^33!+\fracx^44!+\cdots$$Taylor series of $e^ix$:$$\beginaligne^ix&=1+ix+\frac(ix)^22!+\frac(ix)^33!+\frac(ix)^44!\cdots\\&=\colorgreen1+ix\colorgreen-\fracx^22!-\fracix^33!\colorgreen+\fracx^44!\cdots\endalign$$

Taylor series of $e^-ix$:$$\beginaligne^-ix&=1-ix+\frac(-ix)^22!+\frac(-ix)^33!+\frac(-ix)^44!\cdots\\&=\colorgreen1-ix\colorgreen-\fracx^22!+\fracix^33!\colorgreen+\fracx^44!\cdots\endalign$$

So including the above two together, the imaginary parts cancel out each other and also the real components are twice those in $\cos(x)$.

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