# Difference between revisions of "2003 AIME I Problems/Problem 4"

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<math> \log_{10} (\sin x + \cos x) = \frac{1}{2}(\log_{10} n - \log_{10} 10) </math> | <math> \log_{10} (\sin x + \cos x) = \frac{1}{2}(\log_{10} n - \log_{10} 10) </math> | ||

− | <math> \log_{10} (\sin x + \cos x) = \frac{1}{2}(\log_{10} \frac{n}{10}) </math> | + | <math> \log_{10} (\sin x + \cos x) = \frac{1}{2}\left(\log_{10} \frac{n}{10}\right) </math> |

− | <math> \log_{10} (\sin x + \cos x) = (\log_{10} \sqrt{\frac{n}{10}}) </math> | + | <math> \log_{10} (\sin x + \cos x) = \left(\log_{10} \sqrt{\frac{n}{10}}\right) </math> |

<math> \sin x + \cos x = \sqrt{\frac{n}{10}} </math> | <math> \sin x + \cos x = \sqrt{\frac{n}{10}} </math> | ||

− | <math> (\sin x + \cos x)^{2} = (\sqrt{\frac{n}{10}})^2 </math> | + | <math> (\sin x + \cos x)^{2} = \left(\sqrt{\frac{n}{10}}\right)^2 </math> |

<math> \sin^2 x + \cos^2 x +2 \sin x \cos x= \frac{n}{10} </math> | <math> \sin^2 x + \cos^2 x +2 \sin x \cos x= \frac{n}{10} </math> | ||

− | <math> 1 + 2(\frac{1}{10}) = \frac{n}{10} </math> | + | <math> 1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} </math> |

<math> \frac{12}{10} = \frac{n}{10} </math> | <math> \frac{12}{10} = \frac{n}{10} </math> |

## Revision as of 18:34, 4 November 2006

## Problem

Given that and that find

## Solution