1. **State the problem:** We want to prove or verify the identity: $$-(-1)^n\sin(x\pi) = \cos\left(x\pi + (-1)^n \frac{\pi}{2}\right)$$ 2. **Recall important formulas and rules:** - The sine and cosine functions have phase shift relationships. - Euler's formula and trigonometric identities can help. - Key identity: $$\cos(\theta + \frac{\pi}{2}) = -\sin(\theta)$$ and $$\cos(\theta - \frac{\pi}{2}) = \sin(\theta)$$ 3. **Analyze the right side:** Consider the term $$\cos\left(x\pi + (-1)^n \frac{\pi}{2}\right)$$. - If $(-1)^n = 1$ (when $n$ is even), then: $$\cos\left(x\pi + \frac{\pi}{2}\right) = -\sin(x\pi)$$ - If $(-1)^n = -1$ (when $n$ is odd), then: $$\cos\left(x\pi - \frac{\pi}{2}\right) = \sin(x\pi)$$ 4. **Analyze the left side:** $$-(-1)^n \sin(x\pi)$$ - For even $n$, $(-1)^n = 1$, so left side is: $$-1 \cdot \sin(x\pi) = -\sin(x\pi)$$ - For odd $n$, $(-1)^n = -1$, so left side is: $$-(-1) \sin(x\pi) = \sin(x\pi)$$ 5. **Compare both sides:** - For even $n$, both sides equal $-\sin(x\pi)$. - For odd $n$, both sides equal $\sin(x\pi)$. 6. **Conclusion:** The identity holds for all integers $n$ and real $x$: $$-(-1)^n \sin(x\pi) = \cos\left(x\pi + (-1)^n \frac{\pi}{2}\right)$$