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Trigonometry Factoring exponents from trig functions


Jun 27, 2012
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:

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

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?


Well-known member
Feb 2, 2012
Hello, daigo!

[tex]\text{Euler's formula: }\:e^{ix} \:=\:\cos x + i\sin x[/tex]

[tex]\text{Square both sides:}[/tex]
. . [tex](e^{ix})^2 \:=\: (\cos x + i\sin x)^2 \:=\: (\cos^2x - \sin^2x) + i(2\sin x\cos x) [/tex] .[1]

[tex]\text{We know that:}[/tex]
. . [tex](e^{ix})^2 \:=\:e^{2ix} \:=\:e^{i(2x)} \:=\:\cos(2x) + i\sin(2x) [/tex] .[2]

Equate real components and imaginary components of [2] and [1].

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