Prove…

Prove: $\ \cos2n\theta=(-1)^n\displaystyle\prod_{k=1}^n\left[1-\dfrac{\cos^2\theta}{\cos^2\left(\frac{(2k-1)\pi}{4n}\right)}\right]$.

Step 1 · Locate the zeros

$\cos2n\theta=0$ exactly when $2n\theta=\frac\pi2+k\pi$ for integer $k$, i.e.:

$$ \theta_k=\frac{(2k+1)\pi}{4n}, \qquad k=0,1,\ldots,2n-1 $$

giving $2n$ distinct roots in a period, matching the degree of $\cos2n\theta$ as a polynomial in $\cos\theta$.

Step 2 · Pair the roots to get $n$ distinct values of $\cos^2\theta$

Pair $\theta_k$ with $\theta_{2n-1-k}$:

$$ \theta_{2n-1-k}=\frac{(4n-2k-1)\pi}{4n}=\pi-\frac{(2k+1)\pi}{4n}=\pi-\theta_k $$

and since $\cos(\pi-\theta_k)=-\cos\theta_k$, both roots give the same $\cos^2\theta$. Relabeling with $k=1,\ldots,n$ via $2k-1=2(k-1)+1$, the $n$ distinct values are:

$$ s_k=\cos^2\left(\frac{(2k-1)\pi}{4n}\right), \qquad k=1,\ldots,n $$

Step 3 · Factor as a polynomial in $\cos^2\theta$

Since $\cos2n\theta$ vanishes exactly at $\cos^2\theta=s_k$ for $k=1,\ldots,n$:

$$ \cos2n\theta=C\prod_{k=1}^n\left(1-\frac{\cos^2\theta}{s_k}\right) $$

with each factor normalized to equal $1$ where $\cos\theta=0$.

Step 4 · Determine $C$ at $\theta=\pi/2$

At $\theta=\pi/2$: $\cos\theta=0$, so every factor on the right equals $1$, giving $C=\cos(n\pi)=(-1)^n$.

Step 5 · Conclusion

$$ \boxed{\cos2n\theta=(-1)^n\prod_{k=1}^n\left[1-\frac{\cos^2\theta}{\cos^2\left(\frac{(2k-1)\pi}{4n}\right)}\right]} $$

Verification

The general identity was checked symbolically for $n=1,2,3,4$ (difference simplifies to $0$).