# Difference between revisions of "2005 AIME II Problems/Problem 9"

## Problem

For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$?

## Solution

### Solution 1

We know by De Moivre's Theorem that $(\cos t + i \sin t)^n = \cos nt + i \sin nt$ for all real numbers $t$ and all integers $n$. So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem.

Recall the trigonometric identities $\cos (\frac{\pi}2 - u) = \sin u$ and $\sin (\frac{\pi}2 - u) = \cos u$ hold for all real $u$. If our original equation holds for all $t$, it must certainly hold for $t = \frac{\pi}2 - u$. Thus, the question is equivalent to asking for how many positive integers $n \leq 1000$ we have that $\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)$ holds for all real $u$. $\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^n = \cos nu + i\sin nu$. We know that two complex numbers are equal if and only if both their real part and imaginary part are equal. Thus, we need to find all $n$ such that $\cos n u = \sin n\left(\frac\pi2 - u\right)$ and $\sin nu = \cos n\left(\frac\pi2 - u\right)$ hold for all real $u$. $\sin x = \cos y$ if and only if either $x + y = \frac \pi 2 + 2\pi \cdot k$ or $x - y = \frac\pi2 + 2\pi\cdot k$ for some integer $k$. So from the equality of the real parts we need either $nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$, in which case $n = 1 + 4k$, or we need $-nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$, in which case $n$ will depend on $u$ and so the equation will not hold for all real values of $u$. Checking $n = 1 + 4k$ in the equation for the imaginary parts, we see that it works there as well, so exactly those values of $n$ congruent to $1 \pmod 4$ work. There are $\boxed{250}$ of them in the given range.

### Solution 2

This problem begs us to use the familiar identity $e^{it} = \cos(t) + i \sin(t)$. Notice, $\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}$ since $\sin(-t) = -\sin(t)$. Using this, $(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)$ is recast as $(i e^{-it})^n = i e^{-itn}$. Hence we must have $i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}$. Thus since $1000$ is a multiple of $4$ exactly one quarter of the residues are congruent to $1$ hence we have $\boxed{250}$.

### Solution 3

De Moivre's Theorem states that $(\cos(t) + i \sin(t))^n = \cos(nt) + i \sin(nt).$ However, our given expression is $(\sin (t) + i \cos (t))^n$. We use the identity $\sin(x) = \cos(90-x)$ to substitute things in, getting $(\cos(90-t) + i \sin(90-t))^n$. We now use De Moivre's, getting $\cos(90n-nt) + i \sin(90n-nt)$. We want this to be equal to $\cos(90-nt) + i \sin(90-nt)$. This is only true when the two angles are the same, or when their difference is a multiple of $90$. $(90n - nt) - (90 - nt) = 90(n-1)$. $90(n-1)$ is only a multiple of $360$ when $n-1$ is a multiple of $4$, or when $n \equiv 1 \bmod {4}$. There are therefore $\boxed{250}$ possible values of $n$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 