1. **State the problem:** Simplify the expression $$\frac{\sin(16k) + \sin(8k)}{\cos(16k) + \cos(8k)}$$. 2. **Recall sum-to-product formulas:** - For sine: $$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ - For cosine: $$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ 3. **Apply the formulas:** $$\sin(16k) + \sin(8k) = 2 \sin\left(\frac{16k + 8k}{2}\right) \cos\left(\frac{16k - 8k}{2}\right) = 2 \sin(12k) \cos(4k)$$ $$\cos(16k) + \cos(8k) = 2 \cos\left(\frac{16k + 8k}{2}\right) \cos\left(\frac{16k - 8k}{2}\right) = 2 \cos(12k) \cos(4k)$$ 4. **Substitute back into the expression:** $$\frac{2 \sin(12k) \cos(4k)}{2 \cos(12k) \cos(4k)}$$ 5. **Cancel common factors:** $$\frac{\cancel{2} \sin(12k) \cancel{\cos(4k)}}{\cancel{2} \cos(12k) \cancel{\cos(4k)}} = \frac{\sin(12k)}{\cos(12k)}$$ 6. **Recognize the quotient:** $$\frac{\sin(12k)}{\cos(12k)} = \tan(12k)$$ **Final answer:** $$\boxed{\tan(12k)}$$