Question 575481
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Now that I have managed to unstump myself...


First consider the case *[tex \Large x\ =\ 0] in which case the equation reduces to *[tex \Large 0\ =\ 0].


Next consider *[tex \Large x\ \not =\ 0], so *[tex \Large \sin(3x)\ \not =\ 0] allowing:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin^3(3x)}{\sin(3x)}\ =\ \frac{\frac{1}{2}\left(\sin(3x)\right)\left(1\ -\ \cos(6x)\right)}{\sin(3x)}]


which yields:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(3x)\ =\ \frac{1}{2}\left(1\ -\ \cos(6x)\right)]


Still ugly, but an improvement.


Let *[tex \LARGE u\ =\ 3x], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(u)\ =\ \frac{1}{2}\left(1\ -\ \cos(2u)\right)]


And


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sin^2(u)\ =\ 1\ -\ \cos(2u)]


Then use *[tex \LARGE \cos(2\varphi)\ =\ \cos^2(\varphi)\ -\ \sin^2(\varphi)]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sin^2(u)\ =\ 1\ -\ \cos^2(u)\ +\ \sin^2(u)]


Then use *[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(\varphi)\ =\ 1\ -\ \cos^2(\varphi)]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sin^2(u)\ =\ \sin^2(u)\ +\ \sin^2(u)]


and finally


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sin^2(u)\ \equiv\ 2\sin^2(u)]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^3(3x)\ \equiv\ \frac{1}{2}\left(\sin(3x)\right)\left(1\ -\ \cos(6x)\right)]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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