Scott Ripley

Posted on Thursday, 10 July, 2003 - 09:46 am:

I am given that

z^{n} + 1 / z^{n} = 2 \cos n θ

, for

n

a positive integer, and

z = cis θ

. I am supposed to show that

4 \cos^{3} θ = \cos 3 θ + 3 \cos θ

In part (a) of this question I was to show that $(z + 1 / z)^{3} = (z^{3} + 1 / z^{3}) + (3 z + 3 / z)$ which I showed easily. I'm supposed to use the result from (a) to show that $4 \cos^{3} θ = \cos 3 θ + 3 \cos θ$ but I don't know how.

I know that $4 \cos^{3} θ = 2 (2 \cos n θ)$ so that is in my mind but from there I get stuck. I try substituting $cis θ$ into the large equation above but I can't find any things to cancel out to ultimately get just $\cos$ .

Andre Rzym

Posted on Thursday, 10 July, 2003 - 10:10 am:

You are almost there. Firstly, you mean

$z = \cos (θ) + i \sin (θ)$

and we are to prove

$4 \cos^{3} θ = . . .$

Now focus on each term of your identity:

$(z + 1 / z)^{3} = (z^{3} + 1 / z^{3}) + (3 z + 3 / z)$

Now we know

$(z + 1 / z) = 2 \cos (θ)$

$8 \cos^{3} (θ) = (z^{3} + 1 / z^{3}) + (3 z + 3 / z)$

Can you carry on from here?

Andre

Scott Ripley

Posted on Thursday, 10 July, 2003 - 10:37 am:

I understood how you got

8 \cos^{3} (θ) = (z^{3} + 1 / z^{3}) + (3 z + 3 / z)

but from there I'm just stuck again. I tried substituting

cis θ

into

(z^{3} + 1 / z^{3}) + (3 z + 3 / z)

but once again nothing. I thought that if

8 \cos^{3} (θ) = (z^{3} + 1 / z^{3}) + (3 z + 3 / z)

then if I divided it by two it would give me

4 \cos^{3} (θ)

which is what I require, but then I don't know how dividing all the cos and sin parts by 2 would do any good. Is there something I'm missing here?

Andre Rzym

Posted on Thursday, 10 July, 2003 - 11:02 am:

Now focus on the

(3 z + 3 / z)

term. We can rewrite this as

3 (z + 1 / z)

. But from your very first statement, what does

z + 1 / z

equal?

This leaves the $z^{3} + 1 / z^{3}$ term. For this, note that $z^{3} = \cos (3 θ) + i \sin (3 θ)$ . What, therefore, does $1 / z^{3}$ equal? What, therefore, does $z^{3} + 1 / z^{3}$ equal?

Andre

Scott Ripley

Posted on Thursday, 10 July, 2003 - 12:44 pm:

Oh i got it now, thanks a bunch