[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.

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...

are you talking about cos cubed or cos(3x)?

What would be the point in asking how to write cos^3 in terms of cos^2...?

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

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

wow thank you very much for your help that is very useful to me i will be investigating DeMoivre's theorem accordingly

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

oh shoot the question asked to express it as a sum of powers of cos(x) if you were confused by the wording

That's alright then. So...

cos(3x) = = 4 cos^3(x) - 3cos(x)