1961 IMO Problems/Problem 3

Problem

Solve the equation

$\cos^n{x} - \sin^n{x} = 1$

where $n$ is a given positive integer.

Solution 1

Since $cos^2x + sin^2x = 1$ , we cannot have solutions with $n\ne2$ and $0<|cos(x)|,|sin(x)|<1$ . Nor can we have solutions with $n=2$ , because the sign is wrong. So the only solutions have $sin (x) = 0$ or $cos (x) = 0$ , and these are: $x =$ multiple of $\pi$ , and $n$ even; $x$ even multiple of $\pi$ and $n$ odd; $x$ = even multiple of $\pi + \frac{3\pi}{2}$ and $n$ odd.

Solution 2 (Calculus)

First consider $n=1$ . Squaring both sides yields $\cos^2{x}-2\cos{x}\sin{x}+\sin^2{x}=1$ , and utilizing identities yields $\sin{2x}=0$ . This results in solutions $x=\frac{\pi}{2}k$ for integers $k$ . After manually checking the solutions in the original equation, we conclude that only $x=2k\pi$ and $x=\frac{3\pi}{2}+2k\pi$ are solutions to the equation.

Next, we consider $n=2$ . This is equivalent to $\cos{2x}=1$ , so the solutions are $x=k\pi$ for integers $k$ . Once again, substituting into the equation results in a valid identity, so these are the only solutions.

Let $f(x)=\cos^n{x}-\sin^n{x}$ . Taking the derivative yields $f'(x)=-n\sin{x}\cos{x}(\sin^{n-2}{x}+\cos^{n-2}{x})$ for $n>2$ . We wish to consider extrema, so we find all $x$ -values such that $f'(x)=0$ by checking the zeros of each factor. In $x\in[0,2\pi]$ , $f'(x)=0$ for even $n$ when $x=0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi$ , and for odd $n$ when $x=0,\frac{\pi}{2},\frac{3\pi}{4},\pi,\frac{3\pi}{2},\frac{7\pi}{4},2\pi$ . The differences between the two come from the $\sin^{n-2}{x}+\cos^{n-2}{x}$ factor, reaching zeros when $\tan^{n-2}{x}=-1$ . This is impossible when $n$ is even, and if $n$ is odd, it implies that $\tan{x}=-1$ , hence the solutions.

We then calculate the value of $f(x)$ at every such $x$ . For even $n$ , the $x$ -values $0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi$ result in $f(x)$ -values of $1,-1,1,-1,1$ . Since these are the only possible extrema, we conclude that $1$ is always a maximum and $-1$ is always a minimum; thus the only solutions are the maxima, which occur at $x=0,\pi,2\pi$ when $x\in[0,2\pi]$ ; more generally, we have $x=k\pi$ for integers $k$ .

Next, for odd $n$ , the $x$ -values $x=0,\frac{\pi}{2},\frac{3\pi}{4},\pi,\frac{3\pi}{2},\frac{7\pi}{4},2\pi$ yield $f(x)$ -values of $1,-1,-2\left(\frac{1}{\sqrt{2}}\right)^n,-1,1,2\left(\frac{1}{\sqrt{2}}\right)^n,1$ . Obviously $x=0,\frac{3\pi}{2},2\pi$ are all solutions; we show that on our closed domain these are the only solutions. On $x=0$ to $x=\frac{\pi}{2}$ , the function is always decreasing; if it were increasing at any point, an extremum would be required to change the sign of the derivative; however, there does not exist an extremum between these two points (since we have already found all the extrema!); thus only $x=0$ satisfies the equation. We can repeat this argument for all values between two extrema; eventually, we can conclude that the only times that $f(x)=1$ are the above (since no segment contains $1$ at any point other than an endpoint; as a result, the function cannot equal $1$ at any point other than an endpoint of some segment, which are the extrema). As a result, extending this to all $x$ , the solutions are $x=2k\pi$ and $x=\frac{3\pi}{2}+2k\pi$ for integers $k$ .

Since these results match our $n=1$ and $n=2$ cases, we conclude that when $n$ is odd, the solutions are $x=2k\pi$ and $x=\frac{3\pi}{2}+2k\pi$ for integers $k$ , and when $n$ is even, the solutions are $x=k\pi$ for integers $k$ .

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1961 IMO (Problems) • Resources
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4
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1961 IMO Problems/Problem 3

Problem

Solution 1

Solution 2 (Calculus)