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FP3 roots

Having derived the expression for tan(4x) in terms of tan(x), and then found cot(4x) in terms of cot(x), using de Moivre's theorem, how should I go about showing that one of the roots of the equation x2-6x+1=0 is cot2(1/8*pi) ?
Reply 1
Avatar for MPCFM
MPCFM
x^2 - 6x + 1 = 0


Assuming that the question gives you the fact that cot^2(pi/8) is an actual root, and asks you to show it:

then simply plug in cot^2(pi/8)
as x into the equation: x^2-6x+1.
this is equal to 0 therefore it can be said that it is a root.



or you could use completing the square (although this doesnt give you a value in terms of the trigonometric function cot.)


The idea; adding the same value on both sides maintains the equality (the equality remains true).


You look for a value that you can add (or subtract) in order to turn your quadratic into a square:


In this case, if you add 8 on both sides, you can get:


x^2 -6x + 9 = 8


The left side is the same value as:
(x-3)(x-3)
in other words, it is a square
(x-3)^2 = x^2 - 6x + 9


Knowing that, you can now solve the equation by taking the square root on both sides


(x-3)^2 = 8
(x-3) = +/- √8
x = 3 +/- √8


( x has two possible values: 3 + √8 and 3 - √8 )

and if you plug in: 1/(cot^2(pi/8)) = 3 + √8
(edited 9 years ago)
Original post by Big-Daddy
Having derived the expression for tan(4x) in terms of tan(x), and then found cot(4x) in terms of cot(x), using de Moivre's theorem, how should I go about showing that one of the roots of the equation x2-6x+1=0 is cot2(1/8*pi) ?


I assume you have shown that cot⁡4x=cot⁡2x−6cot⁡x+1\cot 4x=\cot^2x-6\cot x+1 so the roots of cot⁡2x−6cot⁡x1=0 \cot ^2x-6\cot x1=0 are also the roots of cot⁡4x=0⇒4x=π2\cot 4x=0 \Rightarrow 4x=\frac{\pi}{2}
Reply 3
Avatar for BabyMaths
BabyMaths
Shouldn't that be cot⁡(4θ)≡cot⁡4(θ)−6cot⁡2(θ)+14cot⁡3(θ)−4cot⁡(θ)\displaystyle \cot(4\theta)\equiv \frac{\cot^4(\theta)-6\cot^2(\theta)+1}{4\cot^3( \theta)-4\cot(\theta)} and then with θ=π2\theta=\frac{\pi}{2} and x=cot⁡2(θ)x=\cot^2(\theta) it works out?
Reply 4
Avatar for Big-Daddy
Big-Daddy
OP
13
Got it, thanks guys.

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