Verifying a Hard Trigonometric Identitiy ()

In summary: I see the difference of squares. Should I factor out a cosx now?No, you don't want to factor out a cos x.
  • #1
Harleyd2900
5
0
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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  • #2
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
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
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
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
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,
 
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What is the purpose of verifying a hard trigonometric identity?

Verifying a hard trigonometric identity is important in mathematics because it allows us to prove that two seemingly different expressions are in fact equal. This helps to establish the validity of mathematical theories and formulas.

What are the steps involved in verifying a hard trigonometric identity?

The first step is to identify the identity that needs to be verified. Then, using known trigonometric identities and properties, manipulate one side of the equation until it is equivalent to the other side. Finally, substitute in different values for the variables to test the equality of the two sides.

Why is verifying a hard trigonometric identity considered difficult?

Verifying a hard trigonometric identity can be difficult because it requires a deep understanding of trigonometric functions and identities, as well as the ability to manipulate equations using these identities. It also often involves complex algebra and multiple steps, making it a time-consuming process.

What are some common mistakes to avoid when verifying a hard trigonometric identity?

Common mistakes when verifying a hard trigonometric identity include forgetting to use the correct trigonometric identities, making errors in algebraic manipulations, and not properly simplifying expressions. It is also important to check the validity of any substitutions made during the process.

How can I check if my solution to a hard trigonometric identity is correct?

The best way to check if your solution to a hard trigonometric identity is correct is to substitute in different values for the variables and compare the two sides of the equation. If they are equal for all values, then the identity is verified. You can also ask a colleague or teacher to review your solution for any errors.

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