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 $n.$

$\log_{10} \sin x + \log_{10} \cos x = -1$

$\log_{10}(\sin x \cos x) = -1$

$\sin x \cos x = \frac{1}{10}$

$\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1)$

$\log_{10} (\sin x + \cos x) = \frac{1}{2}(\log_{10} n - \log_{10} 10)$

$\log_{10} (\sin x + \cos x) = \frac{1}{2}\left(\log_{10} \frac{n}{10}\right)$

$\log_{10} (\sin x + \cos x) = \left(\log_{10} \sqrt{\frac{n}{10}}\right)$

$\sin x + \cos x = \sqrt{\frac{n}{10}}$

$(\sin x + \cos x)^{2} = \left(\sqrt{\frac{n}{10}}\right)^2$

$\sin^2 x + \cos^2 x +2 \sin x \cos x= \frac{n}{10}$

$1 + 2\left(\frac{1}{10}\right) = \frac{n}{10}$

$\frac{12}{10} = \frac{n}{10}$

$n = 12$