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

Revision as of 17:34, 4 November 2006

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 $n.$

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

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

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

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

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

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

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

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

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

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

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

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

$n = 12$

@@ Line 12: / Line 12: @@
 <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)^{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> 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>