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

Revision as of 21:42, 26 October 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 $n.$

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

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

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

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

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

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

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

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

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

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

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

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

$n = 12$

@@ Line 26: / Line 26: @@
 <math> \frac{12}{10} = \frac{n}{10} </math>
-<math> n = 012 </math>
+<math> n = 12 </math>
 == See also ==