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Aug 7th, 2006, 02:22 AM
#1
Thread Starter
New Member
[RESOLVED] Trigonometric identities
I am having a few problems with the concept of using trignometric identities in terms of factorising or writing trigonometric equations in different forms.
write cos3(x) in terms of cos^2.
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Aug 7th, 2006, 05:43 AM
#2
Re: Trigonometric identities
The upshot is: you can write trig functions in terms of other trig functions. That's the concept.
For the details, look here.
So you might write cos(3x) as cos (2x + x) and then use the addition formulae and the double angle formulae...
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Aug 7th, 2006, 05:55 AM
#3
Conquistador
Re: Trigonometric identities
are you talking about cos cubed or cos(3x)?
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Aug 7th, 2006, 06:03 AM
#4
Re: Trigonometric identities
What would be the point in asking how to write cos^3 in terms of cos^2...?
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Aug 7th, 2006, 10:13 AM
#5
Thread Starter
New Member
Re: Trigonometric identities
thank you for your clues however im still not entirely clear how i can express this in terms of cos^2(x) i keep on getting a jumbled expression with some cos^2x terms
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Aug 7th, 2006, 03:52 PM
#6
Member
Re: Trigonometric identities
cos(3x) == cos(2x + x)
cos(2x+x) = cos(2x)cos(x) - sin(2x)(sinx)
But cos(2x) = 2cos²(x) - 1 and sin(2x) = 2sin(x)cos(x) so
cos(2x+x) = [2cos²(x) - 1]cos(x) - [2sin(x)cos(x)]sin(x)
= 2cos^3(x) - cos(x) - 2sin²(x)cos(x)
But sin²(x) = 1 - cos²(x) so
cos(2x+x) = 2cos^3(x) - cos(x) - 2[1-cos²(x)]cos(x)
= 2cos^3(x) - cos(x) - 2cos(x) + 2cos^3(x)
= 4 cos^3(x) - 3cos(x)
I'm not quite sure what they mean by "in terms of cos²(x)", but maybe this will do:
cos(3x) = cos²x [4 cos(x) - 3/cos(x)]
This whole proof can also be acheived very neatly by saying cos(3x) = Re{e^i(3x)} then applying DeMoivre's theorem. Try it yourself!
All the best, Matt
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Aug 7th, 2006, 04:17 PM
#7
Thread Starter
New Member
Re: Trigonometric identities
wow thank you very much for your help that is very useful to me i will be investigating DeMoivre's theorem accordingly
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Aug 7th, 2006, 04:34 PM
#8
Member
Re: Trigonometric identities
You're welcome. In fact it's all done here
http://everything2.com/index.pl?node_id=891877
and also right at the bottom of the wikipedia article here
http://en.wikibooks.org/wiki/Algebra/Complex_numbers
All the best, Mattywoo
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Aug 7th, 2006, 04:42 PM
#9
Thread Starter
New Member
Re: Trigonometric identities
oh shoot the question asked to express it as a sum of powers of cos(x) if you were confused by the wording
Last edited by seostattler; Aug 7th, 2006 at 04:43 PM.
Reason: incorrect info
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Aug 7th, 2006, 05:00 PM
#10
Member
Re: Trigonometric identities
That's alright then. So...
cos(3x) = = 4 cos^3(x) - 3cos(x)
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