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2005 AIME II Problems/Problem 9

Revision as of 20:49, 15 July 2008 by Cognos599 (talk | contribs) (→‎Solution)

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

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 $(\sin(\frac\pi2 - u) + i \cos(\frac\pi 2 - u))^n = \sin n(\frac\pi2 -u) + i\cos n(\frac\pi2 - u)$ holds for all real $u$.

$(\sin(\frac\pi2 - u) + i \cos(\frac\pi 2 - u))^n = (\cos u + i \sin u)^n = \cos nu + i\sin nu$. 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 $\cos n u = \sin n(\frac\pi2 - u)$ and $\sin nu = \cos n(\frac\pi2 - u)$ 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 $nu + n(\frac\pi2 - u) = \frac\pi 2 + 2\pi \cdot k$, in which case $n = 1 + 4k$, or we need $-nu + n(\frac\pi2 - u) = \frac\pi 2 + 2\pi \cdot k$, 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 250 of them in the given range.

Solution 2

We can rewrite $(\sin t + i \cos t)^n$ as $(\cos(90-t) + i\sin(90-t))^n$ which, by De Moivre's Theorem is equal to $\cos{n(90-t)} + i\sin{n(90-t)}$, but we know that is is equal to $\sin nt + i \cos nt$, now if we replace $\sin nt + i \cos nt$ with $\cos{90-nt} + i\sin{90-nt}$. This gives us the equation:


$\cos{90n - nt} + i\sin{90 - nt} = \cos{90-nt} + i\sin{90-nt}$

Equating the real parts or the imaginary parts will give the same solution set, so we will equate the real parts. So we get

$90n - nt = 90 - nt$

but $90$ in the right hand side of the equation is just the principal value, but we can have any equivalent value. So our new equation is:

$90n = 90 + 360x$ for $x$ being an integer.

$n = 4x + 1$

we know that $n \le 1000$

so we want all numbers that are less than or equal to $1000$ and also $\equiv 1 (\mbox{mod} 4)$ and there are $\fbox{250}$ such numbers

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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