sums of powers

ImOk

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Dec 29, 2008
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Happy New Year!!!

I hope I'm welcome :D

Having trouble solving this problem

"Write cos 3x as sums of powers of cos x"

I tried expanding it by cos 3x = cos (2x + x) and applying the identities obtaining cos 2x cos x - sin 2x sin x

But I don't know what to do next. What does it means to write cos 3x as sums of powers of cos x?


Wish you the best of luck!!!
 
Hello, ImOk!

I hope I'm welcome . . . . Welcome aboard!

Write \(\displaystyle \cos3x\) as sums of powers of \(\displaystyle \cos x.\)

You had a great start . . .

\(\displaystyle \cos(3x) \;=\;\cos(2x + x)\)

. . . . . .\(\displaystyle =\qquad\underbrace{\cos2x}\cos x \quad\;\;- \quad\;\;\underbrace{\sin2x}\sin x\)
. . . . . .\(\displaystyle = \;\overbrace{\left(2\cos^2\!x - 1\right)}\cos x - \overbrace{(2\sin x\cos x)}\sin x\)

. . . . . .\(\displaystyle = \;2\cos^3\!x - \cos x \;\;- \;\;2\cos x\underbrace{\sin^2\!x}\)
. . . . . .\(\displaystyle = \;2\cos^3\!x - \cos x - 2\cos x\overbrace{(1 - \cos^2\!x)}\)

. . . . . .\(\displaystyle = \;2\cos^3\!x - \cos x - 2\cos x + 2\cos^3\!x\)

. . . . . .\(\displaystyle = \;4\cos^3\!x - 3\cos x\)

 
ImOk said:
Happy New Year!!!

I hope I'm welcome :D

Having trouble solving this problem

"Write cos 3x as sums of powers of cos x"

I tried expanding it by cos 3x = cos (2x + x) and applying the identities obtaining cos 2x cos x - sin 2x sin x

But I don't know what to do next. What does it means to write cos 3x as sums of powers of cos x?


Wish you the best of luck!!!

An alternative method that will probably be useless to you unless you have done complex numbers is to use DeMoivre's theorem:

\(\displaystyle \cos(3x)=\mathrm{re}\left((\cos(x)+\mathrm{i} \sin(x))^3\right)\)

........... \(\displaystyle =\mathrm{re}\left(\cos^3(x)+3 \cos^2(x) (\mathrm{i} \sin(x))+3\cos(x) (\mathrm{i} \sin(x))^2+ (\mathrm{i} \sin(x))^3\right)\)

............\(\displaystyle = \cos^3(x)-3\sin^2(x)\cos(x)\)
etc...
 
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