2005 AIME II Problem 9

Question

For how many positive integers

n

less than or equal to

1000

is

\begin{aligned} &(\sin t + i \cos t)^n \\ &= \sin nt + i \cos nt \end{aligned}

true for all real

t?

Answer

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.

Problem 9 in other years