Prove that\[4 \sin (x) \sin (\omega x) \sin (\omega^2 x) = - (\sin (2x) + \sin (2 \omega x) + \sin (2 \omega^2 x))\]

\[\] **Notation:** \(\omega\) denotes a primitive cube root of unity.

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## Comments

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TopNewest\[\begin{align} 4 \sin (x) \sin (\omega x) \sin (\omega^2 x) &= 2 \sin (x) \left( \cos (\omega x - \omega^2 x) - \cos (\omega x + \omega^2 x) \right) \\ &= 2 \sin (x) \left( \cos (2 \omega x + x ) - \cos (x) \right) \\ &= 2 \sin (x) \cos (2 \omega x + x) - \sin (2x) \\ &= \sin (x + 2 \omega x + x) + \sin (x - 2 \omega x - x) - \sin (2x) \\ &= \sin (-2 \omega^2 x) + \sin (-2 \omega x) - \sin (2x) \\ &= - \left( \sin (2x) + \sin (2 \omega x) + \sin (2 \omega^2 x) \right) & \mathbf{Q.E.D.} \end{align}\]

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@Ishan Singh I think apart from usual trigonometry, this should mean something. How do you interpret "sines" of complex numbers geometrically?

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Hyperbolic functions, like trigonometric, but on a hyperbola instead of a circle. For z = a+ib, you can use addition formula etc. and then rewrite in terms of hyperbolic trigonometric functions.

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