MHB Factoring exponents from trig functions

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The discussion revolves around deriving double angle identities using Euler's formula, e^{ix} = cos x + i sin x. The participant questions the validity of pulling the '2' from e^{i2x} to square the right-hand expression. The response clarifies that squaring both sides of Euler's formula leads to the identities for cos(2x) and sin(2x) through equating real and imaginary components. The derived identities are cos(2x) = cos^2(x) - sin^2(x) and sin(2x) = 2sin(x)cos(x). This method effectively demonstrates the relationship between exponential and trigonometric functions.
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I tend to forget some of the trigonometric functions and someone showed me how to derive the double angle identities from what I think is Euler's formula:

e^{ix} = \cos x + i\sin x
=
e^{i2x} = \cos 2x + i\sin 2x
=
(e^{ix})^{2} = (\cos x + i\sin x)^{2}

I have a question about this step...I understand how the '2' from the e^(i2x) was pulled out, but is it okay to pull the '2' out from the '2x' of the trigonometric functions and to square the right hand expression? How does that work?
 
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Hello, daigo!\text{Euler's formula: }\:e^{ix} \:=\:\cos x + i\sin x\text{Square both sides:}
. . (e^{ix})^2 \:=\: (\cos x + i\sin x)^2 \:=\: (\cos^2x - \sin^2x) + i(2\sin x\cos x) .[1]\text{We know that:}
. . (e^{ix})^2 \:=\:e^{2ix} \:=\:e^{i(2x)} \:=\:\cos(2x) + i\sin(2x) .[2]Equate real components and imaginary components of [2] and [1].

. . . \begin{Bmatrix}\cos2x &=& \cos^2x - \sin^2x \\ \sin2x &=& 2\sin x\cos x \end{Bmatrix}
 
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