Verifying a Hard Trigonometric Identitiy (please help)

  • #1
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)
 
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Answers and Replies

  • #2
JasonRox
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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.
 
  • #3
Tide
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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]
 
  • #4
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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  • #5
Tide
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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]?
 
  • #6
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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  • #7
HallsofIvy
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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.
 
  • #8
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:
 
  • #9
Whould this bring me anyware:

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


?
 
  • #10
JasonRox
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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
VietDao29
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Hmm,... have you learnt 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,
 
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