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**Hard Prove****Guest**

My Pre-Cal teacher gave us this problem today. I have worked on it for a very long time and have goten no where. I was wondering if anyone had any ideas.

cos²5x - cos²x = -sin4x x sin6x

If that is not clear

cos sqared of five x minus cos squared of x = negative sin of 4x times sin of 6x

Thanks a lot for your help,

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

The title is "Proving an Idenity" so does this mean you are not allowed to use trig identities? As that would be begging the question.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**harleyd2900****Member**- Registered: 2005-12-14
- Posts: 5

You can use the identities. The words he really used were varifying Trigonometric Idenities. "Proving" that one side equals the other side, or the working of one side to equal the other.

Hope that helped.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

http://www.pen.k12.va.us/Div/Winchester/jhhs/math/lessons/trig/ident2.html

sinAsinB = 1/2(cos(A-B) - cos(A+B))

-sinA = sin(-A)

cos(2A) = cos²A - sin²A

-sin(4x) = sin(-4x)

sin(-4x) * sin(6x) = 1/2(cos(-4x - 6x) - cos(-4x + 6x)) = 1/2(cos(-10x) - cos(2x))

1/2(cos(2 * -5x) - cos(2 * x)) = 1/2(cos²(-5x) - sin²(-5x) - ( cos²(x) - sin²(x) ) )

Since sin²(-5x) = 1 - cos²(-5x):

1/2(cos²(-5x) - (1 - cos²(-5x)) - (cos²(x) - (1 - cos²(x)) ) )

1/2( 2cos²(-5x) - 1 -2cos²(x) + 1) = 1/2(2cos²(-5x) - 2cos²(x)) = cos²(-5x) - cos²(x)

And finally, since cos(-A) = cos(A):

cos²(5x) - cos²(x)

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**harleyd2900****Member**- Registered: 2005-12-14
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darn that is great. However he said it could be proven in 4-5 steps by only working the left side.... Not to be picky, but you wouldn't happen to know how do do it like that?

PS: (Sorry I am being picky, I have a test on this type of problem tomarow.)

Thanks so much for your help

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Not a problem. Mostly all you have to do is work backwards:

cos²(5x) - cos²(x) = 1/2(2cos²(5x)-1+1 - (2cos²(x) - 1 + 1) )

= 1/2(cos²(10x) - cos²(2x) )

Since 10x = 6x - (-4x) and 2x = 6x + (-4x):

= sin(6x)sin(-4x)

= -sin(4x)sin(6x)

I normally do a lot more steps than needed, just so others can easily follow along.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**harleyd2900****Member**- Registered: 2005-12-14
- Posts: 5

Its funny he has never shown us that formula sinAsinB = 1/2(cos(A-B) - cos(A+B)). Is it a given?

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**Ricky****Moderator**- Registered: 2005-12-04
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It's a standard trig identity yes.

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**harleyd2900****Member**- Registered: 2005-12-14
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Are there any "more basic" ways you whould attempt doing it. Or without using that specific formula. IDK what I am asking. Maybe he didn't teach us that because its only a Pre-Cal class.

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**harleyd2900****Member**- Registered: 2005-12-14
- Posts: 5

How would you work it from here?

cos²5x - cos²x = (-sin4x)(sin6x)

(cos5x - cosx)(cos5x + cosx)

?

*Last edited by harleyd2900 (2005-12-14 15:37:21)*

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