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

The first equation, $\displaystyle \log_{10} \sin x + \log_{10} \cos x = -1$ , can be combined under the properties of logarithms to $\displaystyle \log_{10}(\sin x \cos x) = -1$ . Therefore, $\sin x \cos x = \frac{1}{10}$ .

Now, manipulate the second equation, $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1)$ .

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

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

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

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

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

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

$\displaystyle \sin ^2 x + \cos ^2 x = 1$ , and we can substitute the value for $\displaystyle \sin x \cos x$ from above.

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

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

Thus, the solution is $n = 012$ .

2003 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2003 AIME I Problems/Problem 4

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