So, in attempting to compute the condition number for the 2-norm of a matrix, I have stumbled upon a problem i can't resolve.

I have the formula

$$ \frac{1-\cos\left(\frac{n}{n+1} \pi\right)}{1-\cos\left(\frac{1}{n+1} \pi\right)}. $$

I arrived at this because the eigenvalues for my matrix are $2\left(1-\cos\left(\frac{p}{n+1} \pi\right)\right)$ for $p=1,..,n$.

The problem comes from the fact that I need to show that this formula somehow ends up being equal to $$\frac{1}{\tan^2\left(\frac{1}{2(n+1)}\pi\right)} $$

I've arrived at

$$ \frac{1-\cos\left(\pi\frac{n}{n+1}\right)}{2\sin^2\left(\frac{\pi}{2(n+1)}\right)} $$

but can't seem to get any further, mostly due to the $n$ in the numerator of the cosine.

Any help would be appreciated.


1 Answer 1


You have the right idea in exploiting the identity $$1 - \cos\theta \;=\; 2 \sin^2 \frac{\theta}{2}$$ to re-write the denominator. Using that on the numerator gives $$1 - \cos\left( \pi \frac{n}{n+1} \right) = 2\sin^2\frac{\pi n}{2(n+1)} \tag{$\star$}$$ Now, simply observe that the arguments of the sines combine very conveniently: $$\frac{\pi n}{2(n+1)} + \frac{\pi}{2(n+1)}= \frac{\pi(n+1)}{2(n+1)} = \frac{\pi}{2}$$ so that $(\star)$ becomes $$2\sin^2\left( \frac{\pi}{2} - \frac{\pi}{2(n+1)} \right) = 2\cos^2\left(\frac{\pi}{2(n+1)}\right)$$ since $\sin(\pi/2-\theta) = \cos\theta$. The result follows. $\square$

  • 1
    $\begingroup$ thanks a ton! I wasn't aware of that last identity. $\endgroup$
    – Sertii
    Oct 6, 2015 at 8:43

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