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Difference between revisions of "1961 IMO Problems/Problem 3"

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==Problem==
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Solve the equation  
 
Solve the equation  
  
 
<math>\cos^n{x} - \sin^n{x} = 1</math>
 
<math>\cos^n{x} - \sin^n{x} = 1</math>
  
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where ''n'' is a given positive integer.
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where <math>n</math> is a given positive integer.
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{{wikify}}
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==Solution 1==
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Since <math>cos^2x + sin^2x = 1</math>, we cannot have solutions with <math>n\ne2</math> and <math>0<|cos(x)|,|sin(x)|<1</math>. Nor can we have solutions with <math>n=2</math>, because the sign is wrong. So the only solutions have <math>sin (x) = 0</math> or <math>cos (x) = 0</math>, and these are: <math>x =</math> multiple of <math>\pi</math>, and <math>n</math> even; <math>x </math> even multiple of <math>\pi</math> and <math>n</math> odd; <math>x</math> = even multiple of <math>\pi + \frac{3\pi}{2}</math> and <math>n</math> odd.
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==Solution 2 (Calculus)==
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First consider <math>n=1</math>. Squaring both sides yields <math>\cos^2{x}-2\cos{x}\sin{x}+\sin^2{x}=1</math>, and utilizing identities yields <math>\sin{2x}=0</math>. This results in solutions <math>x=\frac{\pi}{2}k</math> for integers <math>k</math>. After manually checking the solutions in the original equation, we conclude that only <math>x=2k\pi</math> and <math>x=\frac{3\pi}{2}+2k\pi</math> are solutions to the equation.
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Next, we consider <math>n=2</math>. This is equivalent to <math>\cos{2x}=1</math>, so the solutions are <math>x=k\pi</math> for integers <math>k</math>. Once again, substituting into the equation results in a valid identity, so these are the only solutions.
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Let <math>f(x)=\cos^n{x}-\sin^n{x}</math>. Taking the derivative yields <math>f'(x)=-n\sin{x}\cos{x}(\sin^{n-2}{x}+\cos^{n-2}{x})</math> for <math>n>2</math>. We wish to consider extrema, so we find all <math>x</math>-values such that <math>f'(x)=0</math> by checking the zeros of each factor. In <math>x\in[0,2\pi]</math>, <math>f'(x)=0</math> for even <math>n</math> when <math>x=0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi</math>, and for odd <math>n</math> when <math>x=0,\frac{\pi}{2},\frac{3\pi}{4},\pi,\frac{3\pi}{2},\frac{7\pi}{4},2\pi</math>. The differences between the two come from the <math>\sin^{n-2}{x}+\cos^{n-2}{x}</math> factor, reaching zeros when <math>\tan^{n-2}{x}=-1</math>. This is impossible when <math>n</math> is even, and if <math>n</math> is odd, it implies that <math>\tan{x}=-1</math>, hence the solutions.
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We then calculate the value of <math>f(x)</math> at every such <math>x</math>. For even <math>n</math>, the <math>x</math>-values <math>0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi</math> result in <math>f(x)</math>-values of <math>1,-1,1,-1,1</math>. Since these are the only possible extrema, we conclude that <math>1</math> is always a maximum and <math>-1</math> is always a minimum; thus the only solutions are the maxima, which occur at <math>x=0,\pi,2\pi</math> when <math>x\in[0,2\pi]</math>; more generally, we have <math>x=k\pi</math> for integers <math>k</math>.
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Next, for odd <math>n</math>, the <math>x</math>-values <math>x=0,\frac{\pi}{2},\frac{3\pi}{4},\pi,\frac{3\pi}{2},\frac{7\pi}{4},2\pi</math> yield <math>f(x)</math>-values of <math>1,-1,-2\left(\frac{1}{\sqrt{2}}\right)^n,-1,1,2\left(\frac{1}{\sqrt{2}}\right)^n,1</math>. Obviously <math>x=0,\frac{3\pi}{2},2\pi</math> are all solutions; we show that on our closed domain these are the only solutions. On <math>x=0</math> to <math>x=\frac{\pi}{2}</math>, 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 <math>x=0</math> satisfies the equation. We can repeat this argument for all values between two extrema; eventually, we can conclude that the only times that <math>f(x)=1</math> are the above (since no segment contains <math>1</math> at any point other than an endpoint; as a result, the function cannot equal <math>1</math> at any point other than an endpoint of some segment, which are the extrema). As a result, extending this to all <math>x</math>, the solutions are <math>x=2k\pi</math> and <math>x=\frac{3\pi}{2}+2k\pi</math> for integers <math>k</math>.
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Since these results match our <math>n=1</math> and <math>n=2</math> cases, we conclude that when <math>n</math> is odd, the solutions are <math>x=2k\pi</math> and <math>x=\frac{3\pi}{2}+2k\pi</math> for integers <math>k</math>, and when <math>n</math> is even, the solutions are <math>x=k\pi</math> for integers <math>k</math>.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
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{{IMO box|year=1961|num-b=2|num-a=4}}

Latest revision as of 23:07, 10 July 2025

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 $k$.

~ eevee9406

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