1. **State the problem:** Determine if the statement "If $\angle K$ is an acute angle, then $\cos(K) = \sin(K)$" is always, sometimes, or never true. 2. **Recall the complementary angle identity:** For any angle $\theta$, $\sin(90^\circ - \theta) = \cos(\theta)$. This means sine and cosine of complementary angles are equal. 3. **Analyze the given statement:** The statement says $\cos(K) = \sin(K)$ for an acute angle $K$. This implies $\cos(K) = \sin(K)$. 4. **Use the identity $\sin(K) = \cos(90^\circ - K)$:** So $\cos(K) = \sin(K) = \cos(90^\circ - K)$. 5. **Set the equality:** $\cos(K) = \cos(90^\circ - K)$. 6. **Solve for $K$:** Cosine is equal when angles are equal or supplementary: $$K = 90^\circ - K \quad \text{or} \quad K = 360^\circ - (90^\circ - K)$$ 7. **From the first equation:** $$K = 90^\circ - K$$ Add $K$ to both sides: $$K + K = 90^\circ$$ $$2K = 90^\circ$$ Divide both sides by 2: $$\cancel{2}K / \cancel{2} = 90^\circ / 2$$ $$K = 45^\circ$$ 8. **From the second equation:** $$K = 360^\circ - 90^\circ + K = 270^\circ + K$$ Subtract $K$ from both sides: $$K - K = 270^\circ$$ $$0 = 270^\circ$$ This is false, so discard. 9. **Conclusion:** The equality $\cos(K) = \sin(K)$ holds only when $K = 45^\circ$, which is an acute angle. **Answer:** The statement is **sometimes** true (only at $45^\circ$).