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

Revision as of 19:58, 4 July 2013

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) &= \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} \\

\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

By the Pythagorean identities, $\sin ^2 x + \cos ^2 x = 1$ , and we can substitute the value for $\sin x \cos x$ from $(*)$ . $1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} \Longrightarrow n = \boxed{012}$ .

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 [[Category:Intermediate Algebra Problems]]
 [[Category:Intermediate Trigonometry Problems]]
+{{MAA Notice}}

2003 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2003 AIME I Problems/Problem 4"

Revision as of 19:58, 4 July 2013

Problem

Solution

See also