Sine Omega Identity

Prove that

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

Notation: ω\omega denotes a primitive cube root of unity.

#Geometry

Note by Ishan Singh
4 years, 7 months ago

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Comments

4sin(x)sin(ωx)sin(ω2x)=2sin(x)(cos(ωxω2x)cos(ωx+ω2x))=2sin(x)(cos(2ωx+x)cos(x))=2sin(x)cos(2ωx+x)sin(2x)=sin(x+2ωx+x)+sin(x2ωxx)sin(2x)=sin(2ω2x)+sin(2ωx)sin(2x)=(sin(2x)+sin(2ωx)+sin(2ω2x))Q.E.D.\begin{aligned} 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{aligned}

Tapas Mazumdar - 4 years, 2 months ago

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

Kartik Sharma - 4 years, 2 months ago

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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.

Ishan Singh - 4 years, 2 months ago

I figured it out! In calculus, there's a way to turn functions into infinite sums of exponents of the variable. Stick a complex number in there.

Kent Hitchborn - 2 weeks, 2 days ago
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