Proof

**dwsmith** · April 11th 2010, 10:34 AM

Let t be a fixed number and let $c=cos(t)$ and $s=sin(t)$ and $\mathbf{x}=(c, cs, cs^2,..., cs^{n-1}, s^n)^T$ .

Show that $\mathbf{x}$ is a unit vector.

Hint: $1+s^2+s^4+...+s^{2n-2}=\frac{1-s^{2n}}{1-s^2}$

Not sure how that even helps unless I want to the hint via induction.

**Failure** · April 11th 2010, 10:40 AM

Originally Posted by dwsmith

Let t be a fixed number and let $c=cos(t)$ and $s=sin(t)$ and $\mathbf{x}=(c, cs, cs^2,..., cs^{n-1}, s^n)^T$ .

Show that $\mathbf{x}$ is a unit vector.

Hint: $1+s^2+s^4+...+s^{2n-2}=\frac{1-s^{2n}}{1-s^2}$

Not sure how that even helps unless I want to the hint via induction.

Yes, sure it helps, for the square of the length of that vector is
$c^2+c^2s^2+\cdots+c^2s^{2n-2}+s^{2n}=c^2\frac{1-s^{2n}}{1-s^2}+s^{2n}=1-s^{2n}+s^{2n}=1$
since $1-s^2=c^2$ .

**dwsmith** · April 11th 2010, 10:43 AM

Originally Posted by Failure

Yes, sure it helps, for the square of the length of that vector is
$c^2+c^2s^2+\cdots+c^2s^{2n-2}+s^{2n}=c^2\frac{1-s^{2n}}{1-s^2}+s^{2n}=1-s^{2n}+s^{2n}=1$
since $1-s^2=c^2$ .

How did you get $c^2$ ?

**Failure** · April 11th 2010, 10:51 AM

Originally Posted by dwsmith

How did you get $c^2$ ?

Sorry, but I don't quite understand your question.
I just wrote down what, imho, the square of the length of that vector $(c,cs,cs^2,\ldots,cs^{n-1},s^n)$ happens do be: namely the sum of the squares of its coordinates. Hence I get
$c^2+(cs)^2+\cdots+(cs^{n-1})^2+(s^n)^2=c^2+c^2s^2+\cdots+c^2s^{2n-2}+s^{2n}$
Can you be a bit more specific as to where, exactly the occurrence of $c^2$ in what I wrote surprises you?

**HallsofIvy** · April 12th 2010, 04:25 AM

If you are referring to his " $1- s^2= c^2$ , remember that you said "let c= cos(t) and s= sin(t)". Failure (he really should change that user name!) is just using the fact that $sin^2(t)+ cos^2(t)= 1$ .

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