How do I prove the identity sin(4x) = 4sin(x)cos^3(x)-4sin^3(x)cos(x)?

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Discussion Overview

The discussion revolves around proving the trigonometric identity sin(4x) = 4sin(x)cos^3(x) - 4sin^3(x)cos(x). Participants explore various approaches to demonstrate this identity, including the use of known trigonometric identities and complex numbers.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests using the double angle formula for sine, specifically sin(2(2x)), to approach the proof.
  • Another participant hints at factoring the right-hand side of the identity to identify necessary identities.
  • A different perspective introduces the use of complex numbers to prove trigonometric identities, mentioning the expressions for sine and cosine in terms of exponentials.
  • Another participant proposes using De Moivre's theorem to expand (cos(4x) + i sin(4x)) and compare imaginary parts to derive the identity.
  • A participant attempts to manipulate the identity directly, substituting 2x with X and applying known identities step-by-step, arriving at the original expression.

Areas of Agreement / Disagreement

Participants present multiple approaches to proving the identity, but there is no consensus on a single method or resolution of the proof. The discussion remains open with various techniques being explored.

Contextual Notes

Some participants rely on specific trigonometric identities and transformations, while others introduce complex numbers, indicating a range of mathematical tools being considered. The discussion does not resolve the proof or confirm any method as definitive.

Slicktacker
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Hi all, I have a question proving an identity:

[tex]sin(4x) = 4sin(x)cos^3(x)-4sin^3(x)cos(x)[/tex]

I can't seem to figure it out. I know I should be using the known identities:

[tex]sin(2x) = 2sin(x)cos(x)[/tex]

and probably:

[tex]cos(2x) = 1-2sin^2(x)[/tex]

but I'm stuck. Please help!

Thanks!
 
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First hint, apply your double angle formula for sin to:

[tex]sin(2(2x))[/tex]

what do you get?

Second hint, try factoring the right hand side to help see what identities you'll need:

[tex]sin(4x) = 4sin(x)cos(x)(cos^2(x)-sin^2(x))[/tex]

How does this compare to your answer above? Can you see where the factors come from?
 
Mathematicians never memorize trig identities. Trig identities are proven using complex numbers. You can if you wish try to prove your identity with complex numbers, you may find it may be easier. If I can remember correctly cos(x)=(e^(ix) +e^(-ix))/2 and sin(x) = (e^(ix))-e^(-ix))/2i (someone correct me if I am wrong). From there all you have to do is algebra instead of working with identities to get the left side= to the right side.
 
Rather one may also use D'Moivre,
(cos4x+isin4x) = (cosx+isinx)^4
expand RHS and compare imaginary parts ... as a bonus u also get cos4x :P
 
let me try
sin (4x) = 4 sin x cos *3 x - 4 sin *3 ( x ) cos ( x)

let 2x = X
sin 2X = 2 sin X cos X
=2 sin 2x cos 2x
=2 [( 2sinx cosx ) ( cos*2 x - sin *2 x ) ]
=2 [ 2sinxcos*3 x - 2sin*3cos x ]
= 4sinx cos*3 x - 4sin*3 x cos x
 

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