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2003 AIME I Problems/Problem 4

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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}\ (*).

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) &=...

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 • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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