سوال ۳۱ کنکور ارشد مهندسی کامپیوتر ۱۴۰۱

Convert both sides to exponential (Euler) form. Since $\cos x = \sin\left(\tfrac{\pi}{2}-x\right)$ and $\sin x = \cos\left(\tfrac{\pi}{2}-x\right)$,

$\sin\theta + i\cos\theta = \cos\!\left(\tfrac{\pi}{2}-\theta\right) + i\sin\!\left(\tfrac{\pi}{2}-\theta\right) = e^{\,i\left(\frac{\pi}{2}-\theta\right)}.$

By De Moivre's theorem,

$\left(\sin\theta + i\cos\theta\right)^{n} = e^{\,i\,n\left(\frac{\pi}{2}-\theta\right)} = \cos\!\left(n\left(\tfrac{\pi}{2}-\theta\right)\right) + i\sin\!\left(n\left(\tfrac{\pi}{2}-\theta\right)\right).$

The left-hand side is

$\sin(n\theta) + i\cos(n\theta) = \cos\!\left(\tfrac{\pi}{2}-n\theta\right) + i\sin\!\left(\tfrac{\pi}{2}-n\theta\right) = e^{\,i\left(\frac{\pi}{2}-n\theta\right)}.$

For the identity to hold (for all $\theta$), the two exponentials must be equal:

$n\left(\tfrac{\pi}{2}-\theta\right) = \left(\tfrac{\pi}{2}-n\theta\right) + 2k\pi, \qquad k\in\mathbb{Z}.$

The $-n\theta$ terms cancel, leaving a condition free of $\theta$:

$\frac{n\pi}{2} = \frac{\pi}{2} + 2k\pi \;\Longrightarrow\; n = 1 + 4k.$

So the equality holds exactly when $n \equiv 1 \pmod{4}$.

Now count $n \in \{1,2,\dots,1001\}$ with $n \equiv 1 \pmod 4$:

$n \in \{1, 5, 9, \dots, 1001\}, \qquad 1001 = 1 + 4\cdot 250.$

Thus $k$ runs from $0$ to $250$, giving $250 - 0 + 1 = 251$ values. The answer is 251 (option 3).

Exam trick: $\sin\theta + i\cos\theta$ is NOT in standard polar form $\cos+i\sin$. Rewrite it as $e^{i(\pi/2-\theta)}$ first; then De Moivre forces $n\equiv 1\pmod 4$, and the count is just $\lfloor (1001-1)/4\rfloor + 1 = 251$.