Verifying a Hard Trigonometric Identitiy ()

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Discussion Overview

The discussion revolves around verifying a trigonometric identity involving the expression cos²5x - cos²x = -sin4x × sin6x. Participants explore various approaches to simplify and prove the identity, sharing hints and techniques related to trigonometric identities and algebraic manipulation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses difficulty in starting the problem and mentions using Pythagorean identities and the difference of squares.
  • Another participant questions the validity of the identity, noting that the left side remains bounded while the right side grows unbounded as x approaches infinity.
  • A hint is provided to rewrite the left side using the difference of squares formula.
  • Participants discuss the application of trigonometric identities involving cos(a - b) and cos(a + b) to the expression.
  • One participant presents a detailed expansion of cos(5x) using angle addition formulas, but another expresses confusion over the complexity of the derivation.
  • Another participant suggests rewriting the right-hand side and expanding it to facilitate simplification.
  • Several identities are proposed to assist in the proof, including sum-to-product identities and double-angle formulas.
  • A participant summarizes a step-by-step approach to reach the left-hand side from the right-hand side, indicating a potential pathway to the solution.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the identity or the best approach to prove it. Multiple competing views and methods are presented, and the discussion remains unresolved.

Contextual Notes

Some participants reference specific trigonometric identities and formulas, but there is no agreement on the application or correctness of these identities in the context of the problem. The discussion includes various assumptions and steps that are not fully resolved.

Harleyd2900
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My Pre-Cal teacher gave us this problem today. I have worked on it for a very long time and have goten no where :confused:. I was wondering if anyone had any ideas on how to do it, or even where to start. I started it myself with using the pythagorean idenities for the left side, then the difference of square; however after that I didn't know where to go.

cos²5x - cos²x = -sin4x x sin6x

Thanks a lot for your help,

Edit: (Damit... sorry I think I posted this in the wrong place)
 
Last edited:
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Harleyd2900 said:
My Pre-Cal teacher gave us this problem today. I have worked on it for a very long time and have goten no where :confused:. I was wondering if anyone had any ideas on how to do it, or even where to start. I started it myself with using the pythagorean idenities for the left side, then the difference of square; however after that I didn't know where to go.
cos²5x - cos²x = -sin4x x sin6x
Thanks a lot for your help,
Edit: (Damit... sorry I think I posted this in the wrong place)

Are you sure that's true?

It's kind of impossible. Because as x goes to infinity the left side remains contained in a small range, where as the right side blows up because you are multiplying by x.
 
Jason,

I think that is a multiplication sign and not the variable x.

Harley,

HINT: Write the left side as

[tex](\cos 5x - \cos x)(\cos 5x + \cos x)[/tex]
 
Tide said:
Jason,
I think that is a multiplication sign and not the variable x.
Harley,
HINT: Write the left side as
[tex](\cos 5x - \cos x)(\cos 5x + \cos x)[/tex]
Good hint. I see the difference of squares. Should I factor out a cosx now?

BTW that x is a multiplication sign... sorry about that.
 
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No, you don't want to factor out a cos x.

Do you know any trig identities involving [itex]\cos a - b[/itex] and [itex]\cos a + b[/itex]?
 
Yes I know both of thoes. Thoes can be used on [tex](\cos 5x - \cos x)(\cos 5x + \cos x)[/tex] ? Whouldn't a cosx have to be on the outside of the () in order to use it?
 
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cos(5x)= cos(4x+ x)= cos(4x)cos(x)- sin(4x)sin(x)

cos(4x)= cos(3x+ x)= cos(3x)cos(x)- sin(3x)sin(x) and
sin(4x)= sin(3x+ x)= sin(3x)cos(x)+ cos(3x)sin(x) so
cos(5x)= (cos(3x)cos(x)- sin(3x)sin(x))cos(x)- (sin(3x)cos(x)+ cos(3x)sin(x))sin(x)
= cos(3x)cos2(x)- sin(3x)sin(x)cos(x)- sin(3x)sin(x)cos(x)- cos(3x)sin2(x)= cos(3x)(cos2x- sin2(x))

etc.
 
HallsofIvy said:
cos(5x)= cos(4x+ x)= cos(4x)cos(x)- sin(4x)sin(x)
cos(4x)= cos(3x+ x)= cos(3x)cos(x)- sin(3x)sin(x) and
sin(4x)= sin(3x+ x)= sin(3x)cos(x)+ cos(3x)sin(x) so
cos(5x)= (cos(3x)cos(x)- sin(3x)sin(x))cos(x)- (sin(3x)cos(x)+ cos(3x)sin(x))sin(x)
= cos(3x)cos2(x)- sin(3x)sin(x)cos(x)- sin(3x)sin(x)cos(x)- cos(3x)sin2(x)= cos(3x)(cos2x- sin2(x))
etc.

You lost me.:eek: :confused:
 
Whould this bring me anyware:

cos²5x - cos²x = (-sin4x)(sin6x)
(1-sin^2(5x)) - (1-sin^2x) =


?
 
  • #10
Harleyd2900 said:
Whould this bring me anyware:
cos²5x - cos²x = (-sin4x)(sin6x)
(1-sin^2(5x)) - (1-sin^2x) =
?

It's basically the answer minus two-three steps.

I'd recommend to go through it all step-by-step. Write them all if you have to.
 
  • #11
Hmm,... have you learned the following identities:
[tex]\cos \alpha + \cos \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right)[/tex]
[tex]\cos \alpha - \cos \beta = -2 \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)[/tex]
[tex]\sin \alpha + \sin \beta = 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right)[/tex]
[tex]\sin \alpha - \sin \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)[/tex]?
---------------------
Back to your problem, we have:
[tex]\cos ^ 2 (5x) - \cos ^ 2 x = (\cos (5x) - \cos x)(\cos (5x) + \cos x)[/tex]. Now with the 4 identities above, and the double-angle formulas, can you prove that [tex]\cos ^ 2 (5x) - \cos ^ 2 x = - \sin (4x) \sin (6x)[/tex]?
Can you go from here?
 
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  • #12
The key is to change the right hand side:

cos²5x - cos²x = -sin6x.sin4x

So cos²5x - cos²x = -sin(5x+x).sin(5x-x)

then expand and simplify the rhs.

The rest is:

- sin4x.sin6x = - (sin5x - x).(sin(5x + x)

= - [(sin5x.cosx - cos5x.sinx).(sin5x.cosx - cos5x.sinx)]

Tip #3, Difference of two squares

= - (sin²5x.cos²x - cos²5x.sin²x)

= - [(1 - cos²5x).cos²x - cos²5x.(1 - cos²x)]

= - [cos²x - cos²5x.cos²x - cos²5x + cos²5x.cos²x]

= - (cos²x - cos²5x)

= cos²5x - cos²x

= LHS

Q.E.D.

The tip numbers refer to trig proof tips in an article I have written,

http://math.suite101.com/article.cfm/trigonometric-identity-advanced-example

Good luck,
 
Last edited by a moderator:

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