# 2003 AIME I Problems/Problem 4

## Problem

Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1),$ find $n.$

## Solution

Using the properties of logarithms, we can simplify the first equation to $\log_{10} \sin x + \log_{10} \cos x = \log_{10}(\sin x \cos x) = -1$. Therefore, $$\sin x \cos x = \frac{1}{10}.\qquad (*)$$

Now, manipulate the second equation. \begin{align*} \log_{10} (\sin x + \cos x) &= \frac{1}{2}(\log_{10} n - \log_{10} 10) \\ \log_{10} (\sin x + \cos x) &= \left(\log_{10} \sqrt{\frac{n}{10}}\right) \\ \sin x + \cos x &= \sqrt{\frac{n}{10}} \\ (\sin x + \cos x)^{2} &= \left(\sqrt{\frac{n}{10}}\right)^2 \\ \sin^2 x + \cos^2 x +2 \sin x \cos x &= \frac{n}{10} \\ \end{align*}

By the Pythagorean identities, $\sin ^2 x + \cos ^2 x = 1$, and we can substitute the value for $\sin x \cos x$ from $(*)$. $1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} \Longrightarrow n = \boxed{012}$.

## See also

 2003 AIME I (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS