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 August 26th, 2009, 10:39 AM #1 Newbie   Joined: Aug 2009 Posts: 2 Thanks: 0 De Moivre's formula and Newton's binomial formula Using De Moivre's formula and Newton's binomial formula show that: a) $\cos 4x= 8 \cos^4 x - 8 \cos^2 x + 1$ b) $\sin 4x= 4 \sin x \cos x (1 - 2 \sin^2 x)$ I really have no idea how to transform them using these formulas??
August 26th, 2009, 03:50 PM   #2
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Re: De Moivre's formula and Newton's binomial formula

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 Originally Posted by agro Using De Moivre's formula and Newton's binomial formula show that: a) $\cos 4x= 8 \cos^4 x - 8 \cos^2 x + 1$ b) $\sin 4x= 4 \sin x \cos x (1 - 2 \sin^2 x)$ I really have no idea how to transform them using these formulas??
It is straigtfoward but tediuous. I'll do the first. See if you can do the second. The main point is using the double angle formulas for sin and cos.
$\cos 4x= 2\cos^2 2x -1 = 2\(2\cos^2 x - 1)^2 - 1 = 8 \cos^4 x - 8 \cos^2 x + 1$

 August 26th, 2009, 04:42 PM #3 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: De Moivre's formula and Newton's binomial formula I think the idea here is to use De Moivre's formula, namely $\cos(nx)+i\sin(nx)=(\cos x+i\sin x)^n.$ Note that if $A=B,$ then $\text{Re}(A)=\text{Re}(B),$ so De M.'s formula can be modified to $\cos(nx)=\text{Re}\left((\cos x+i\sin x)^n\right).$ Therefore, \begin{align}\cos(4x)&=\text{Re}\left(\cos^4x+4i\c os^3x\sin x-6\cos^2x\sin^2x-4i\cos x\sin^3x+\sin^4x\right)\\ &=\cos^4x-6\cos^2x\sin^2x+\sin^4x\\ &=\cos^4x-6\cos^2x(1-\cos^2x)+(1-\cos^2x)^2.\end{align} Expand out to get the required result.
 August 27th, 2009, 06:17 AM #4 Newbie   Joined: Aug 2009 Posts: 2 Thanks: 0 Re: De Moivre's formula and Newton's binomial formula Thanks! I managed to do the second part myself.

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