سوال ۲۶ کنکور ارشد مهندسی کامپیوتر ۱۴۰۲

Rewrite the base using a complementary-angle conversion to standard polar form:

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

By De Moivre's theorem,

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

The left-hand side likewise equals

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

The identity must hold for every $\theta$, so the two arguments must agree modulo $2\pi$:

$\tfrac{\pi}{2} - n\theta \equiv n\tfrac{\pi}{2} - n\theta \pmod{2\pi} \;\Longrightarrow\; \tfrac{\pi}{2} \equiv n\tfrac{\pi}{2} \pmod{2\pi}.$

Dividing by $\tfrac{\pi}{2}$ gives $1 \equiv n \pmod 4$, i.e. $n \equiv 1 \pmod 4$.

Now count the natural numbers $n = 4k + 1 \le 3500$:

$4k + 1 \le 3500 \;\Longrightarrow\; k \le \tfrac{3499}{4} = 874.75 \;\Longrightarrow\; k \in \{0, 1, \dots, 874\}.$

That is $875$ values (from $n = 1$ up to $n = 3497$).

Exam trick: $\sin\theta + i\cos\theta$ is NOT in polar form $\cos + i\sin$. Convert it to $e^{i(\pi/2 - \theta)}$ first; the sign/role swap of sine and cosine is exactly what produces the $n \equiv 1 \pmod 4$ condition rather than the trivial "all $n$".