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 $\cos \left(\frac{\pi}2 - u\right) = \sin u$ and $\sin \left(\frac{\pi}2 - u\right) = \cos u$ 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

We can rewrite $\sin(t)$ as $\cos\left(\frac{\pi}{2}-t\right)$ and $\cos(t)$ as $\sin\left(\frac{\pi}{2}-t\right)$ . This means that $\sin t + i\cos t = e^{i\left(\frac{\pi}{2}-t\right)}=\frac{e^{\frac{\pi i}{2}}}{e^{it}}$ . This theorem also tells us that $e^{\frac{\pi i}{2}}=i$ , so $\sin t + i\cos t = \frac{i}{e^{it}}$ . By the same line of reasoning, we have $\sin nt + i\cos nt = \frac{i}{e^{int}}$ .

For the statement in the question to be true, we must have $\left(\frac{i}{e^{it}}\right)^n=\frac{i}{e^{int}}$ . The left hand side simplifies to $\frac{i^n}{e^{int}}$ . We cancel the denominators and find that the only thing that needs to be true is that $i^n=i$ . This is true if $n\equiv1\pmod{4}$ , and there are $\boxed{250}$ such numbers between $1$ and $1000$ . Solution by Zeroman

Solution 4

We are using degrees in this solution instead of radians. I just process stuff better that way.

We can see that the LHS is $cis(n(90^{\circ}-t))$ , and the RHS is $cis(90^{\circ}-nt)$ So, $n(90-t) \equiv 90-nt \mod 360$ Expanding and canceling the nt terms, we will get $90n \equiv 90 \mod 360$ . Canceling gets $n \equiv 1 \mod 4$ , and thus there are $\boxed{250}$ values of n.

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

2005 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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