×

# 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: $$\omega$$ denotes a primitive cube root of unity.

Note by Ishan Singh
1 year, 2 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

• bulleted
• list

1. numbered
2. list

1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

\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}

- 9 months, 2 weeks ago

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

- 9 months, 2 weeks ago

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.

- 9 months, 2 weeks ago