1. **Problem statement:** Prove the identity for $m \in \mathbb{N}$: $$\cos(2m\theta) = \sum_{k=0}^m \binom{2m}{2k} (-1)^k \cos^{2m-2k}(\theta) \sin^{2k}(\theta)$$ 2. **Recall the double-angle and binomial expansion formulas:** - The double-angle formula for cosine is: $$\cos(2\alpha) = \cos^2(\alpha) - \sin^2(\alpha)$$ - The binomial theorem states: $$ (x + y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k $$ 3. **Express $\cos(2m\theta)$ using Euler's formula:** Recall Euler's formula: $$ e^{i\phi} = \cos(\phi) + i \sin(\phi) $$ Then: $$ \cos(2m\theta) = \text{Re}(e^{i2m\theta}) = \text{Re}((e^{i\theta})^{2m}) $$ 4. **Rewrite $e^{i\theta}$ as $\cos\theta + i\sin\theta$ and expand:** $$ (\cos\theta + i\sin\theta)^{2m} = \sum_{j=0}^{2m} \binom{2m}{j} \cos^{2m-j}(\theta) (i\sin\theta)^j = \sum_{j=0}^{2m} \binom{2m}{j} i^j \cos^{2m-j}(\theta) \sin^j(\theta) $$ 5. **Take the real part to get $\cos(2m\theta)$:** The real part comes from terms where $j$ is even because $i^{2k} = (-1)^k$ is real. So set $j=2k$: $$ \cos(2m\theta) = \sum_{k=0}^m \binom{2m}{2k} (-1)^k \cos^{2m-2k}(\theta) \sin^{2k}(\theta) $$ 6. **Conclusion:** We have derived the given identity by expanding $\cos(2m\theta)$ using Euler's formula and binomial expansion, confirming the identity holds for all natural numbers $m$. **Final answer:** $$\cos(2m\theta) = \sum_{k=0}^m \binom{2m}{2k} (-1)^k \cos^{2m-2k}(\theta) \sin^{2k}(\theta)$$