2005 AIME II 第 9 题

Write $\sin t + i\cos t = i(\cos t - i\sin t)$ and $\sin nt + i\cos nt$ $= i(\cos nt - i\sin nt)$

By de Moivre the equation reduces to $i^n = i,$ so $n$ must be $1$ more than a multiple of $4$

Since $\sin t + i\cos t = i(\cos t - i \sin t)$ and $\sin nt + i\cos nt$ $= i(\cos nt - i\sin nt),$ de Moivre’s theorem (applied to angle $-t$) gives $$\begin{aligned} &(\sin t + i \cos t)^n \\ &= i^n(\cos t - i\sin t)^n \\ &= i^n(\cos nt - i \sin nt). \end{aligned}$$

So the equation holds for all real $t$ exactly when $i^n = i,$ that is, when $n \equiv 1 \pmod 4.$ The values $n = 1, 5, 9, \ldots, 997$ give exactly $250$ positive integers up to $1000.$