Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

FP2 de Moivre's theorem

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202008/6676_01_que_20080620.pdf

Q.6, part c) could anyone explain what they've done in the mark scheme?

MS: http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202008/6676_01_rms_20080807.pdf

Original post by thorn0123

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202008/6676_01_que_20080620.pdf

Q.6, part c) could anyone explain what they've done in the mark scheme?

MS: http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202008/6676_01_rms_20080807.pdf

Let

x^4 - 5x^2 +5 = 0

and let

x= 2\cos\theta

Then

(2\cos\theta)^4 - 5(2\cos\theta)^2 + 5 = 16\cos^4\theta - 20\cos^2\theta + 5 = 0

But we have:

\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta +5\cos\theta \Rightarrow \frac{\cos 5\theta}{\cos\theta} = 16\cos^4\theta - 20\cos^2\theta + 5

Hence

\frac{\cos 5\theta}{\cos\theta} = 0 \Rightarrow \cos 5\theta = 0 \Rightarrow 5\theta = n\frac{\pi}{2}

for

n \in \mathbb{N}

so

\theta = \frac{n\pi}{10}

for

n \in \mathbb{N}

.

Hence

x = 2\cos \frac{n\pi}{10}

are the roots of the equation and in particular,

x = 2\cos \frac{\pi}{10}

is a root.

OP

Original post by atsruser

Let

x^4 - 5x^2 +5 = 0

and let

x= 2\cos\theta

Then

(2\cos\theta)^4 - 5(2\cos\theta)^2 + 5 = 16\cos^4\theta - 20\cos^2\theta + 5 = 0

But we have:

\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta +5\cos\theta \Rightarrow \frac{\cos 5\theta}{\cos\theta} = 16\cos^4\theta - 20\cos^2\theta + 5

Hence

\frac{\cos 5\theta}{\cos\theta} = 0 \Rightarrow \cos 5\theta = 0 \Rightarrow 5\theta = n\frac{\pi}{2}

for

n \in \mathbb{N}

so

\theta = \frac{n\pi}{10}

for

n \in \mathbb{N}

.

Hence

x = 2\cos \frac{n\pi}{10}

are the roots of the equation and in particular,

x = 2\cos \frac{\pi}{10}

is a root.

thanks!

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