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

 
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<math>\cos^n{x} - \sin^n{x} = 1</math>
 
<math>\cos^n{x} - \sin^n{x} = 1</math>
  
where ''n'' is a given positive integer.
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where <math>n</math> is a given positive integer.
  
  
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==Solution==
 
==Solution==
  
{{solution}}
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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.
  
==See Also==
 
  
[[1961 IMO Problems]]
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{{IMO box|year=1961|num-b=2|num-a=4}}

Latest revision as of 01:07, 29 December 2007

Problem

Solve the equation

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

where $n$ is a given positive integer.


Solution

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.


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