Polynomial Question (1 Viewer)

[QZP]

Use the sum and product of roots to evaluate:
i. cos 2pi/9 + cos 4pi/9 - cos pi/9

This question doesn't make much sense to me as it doesn't look like a traditional polynomial.

Give me a hint please (direction)

 

[Trebla]

Use the trig expansion for cos 3A and let x = cos A. When you solve cos 3A = 0 you end up finding the roots of a cubic polynomial. Once you find those roots you can then use sum of roots (with some manipulation of trig identities) to get the result.

 

[QZP]

Use the trig expansion for cos 3A and let x = cos A. When you solve cos 3A = 0 you end up finding the roots of a cubic polynomial. Once you find those roots you can then use sum of roots (with some manipulation of trig identities) to get the result.

But I don't see how cos 3A has anything to do with the question :S

 

[seanieg89]

Use the sum and product of roots to evaluate:
i. cos 2pi/9 + cos 4pi/9 - cos pi/9

This question doesn't make much sense to me as it doesn't look like a traditional polynomial.

Give me a hint please (direction)

Let S be this expression, use that cos(pi/9) = -cos(8pi/9), cos(6pi/9)=-1/2, and cos(pi+x)=cos(pi-x) to express the real part of:


in terms of S.


What familiar polynomial is this the sum of roots of? z^9-1=0.

This gives us a very simple equation to solve for S.

 

[seanieg89]

But I don't see how cos 3A has anything to do with the question :S

I don't see it either, though you could possibly make something like that work.

 

[Carrotsticks]

I don't see it either, though you could possibly make something like that work.

I think Trebla meant cos(3A) = 1/2, as one of the solutions is pi/9.

 

[seanieg89]

I think Trebla meant cos(3A) = 1/2, as one of the solutions is pi/9.

Yeah, I assumed he meant for that to be 1/2 or -1/2. I didn't see the full progression of that working in my head but it seemed like it could work, although I think a bit slower than the method I posted. Would be interested in seeing if there was something quick he had in mind that I was missing.

 

[Drongoski]

Anyway value = 0

 

[anomalousdecay]

Let S be this expression, use that cos(pi/9) = -cos(8pi/9), cos(6pi/9)=-1/2, and cos(pi+x)=cos(pi-x) to express:


in terms of S.


What familiar polynomial is this the sum of roots of? z^9-1=0.

This gives us a very simple equation to solve for S.

OP is referring to 3-unit work not 4-unit. However, your method gives a very neat and quick solution.

 

[Trebla]

Yeah my bad, basically we know that or can easily show that

cos 3A = 4cos³A - 3cos A

Suppose that cos 3A = -1/2 which gives us particular values of A and let x = cos A

The expression becomes the polynomial equation

8x³ - 6x + 1 = 0

When you solve cos 3A = -1/2 you get A = 2pi/9, 4pi/9, 8pi/9

In other words the solutions to the above polynomial are cos 2pi/9, cos 4pi/9, cos 8pi/9

However, note that cos 8pi/9 = -cos pi/9 and now you can easily apply the sum of roots identity to get the result