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Difference between revisions of "2003 AIME I Problems/Problem 4"
m (→‎Solution)
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Given that <math> \log_{10} \sin x + \log_{10} \cos x = -1 </math> and that <math> \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1), </math> find <math> n. </math>
 
Given that <math> \log_{10} \sin x + \log_{10} \cos x = -1 </math> and that <math> \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1), </math> find <math> n. </math>
 
== Solution ==
 
== Solution ==
−
Using the properties of [[logarithm]]s, we can simplify the first equation to <math>\log_{10} \sin x + \log_{10} \cos x = \log_{10}(\sin x \cos x) = -1 </math>. Therefore, <math> \sin x \cos x = \frac{1}{10}\ (*)</math>.
+
Using the properties of [[logarithm]]s, we can simplify the first equation to <math>\log_{10} \sin x + \log_{10} \cos x = \log_{10}(\sin x \cos x) = -1 </math>. Therefore, <cmath> \sin x \cos x = \frac{1}{10}.\qquad (*)</cmath>
  
 
Now, manipulate the second equation.
 
Now, manipulate the second equation.
−
<center><math>\begin{align*}
+
<cmath>\begin{align*}
 
\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} n - \log_{10} 10) \\
 
\log_{10} (\sin x + \cos x) &= \left(\log_{10} \sqrt{\frac{n}{10}}\right) \\
 
\log_{10} (\sin x + \cos x) &= \left(\log_{10} \sqrt{\frac{n}{10}}\right) \\
Line 12: Line 12:
 
\sin^2 x + \cos^2 x +2 \sin x \cos x &= \frac{n}{10} \\
 
\sin^2 x + \cos^2 x +2 \sin x \cos x &= \frac{n}{10} \\
 
\end{align*}
 
\end{align*}
−
</math></center>
+
</cmath>
  
 
By the Pythagorean identities, <math>\sin ^2 x + \cos ^2 x = 1</math>, and we can substitute the value for <math>\sin x \cos x</math> from <math>(*)</math>. <math>1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} \Longrightarrow n = \boxed{012} </math>.
 
By the Pythagorean identities, <math>\sin ^2 x + \cos ^2 x = 1</math>, and we can substitute the value for <math>\sin x \cos x</math> from <math>(*)</math>. <math>1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} \Longrightarrow n = \boxed{012} </math>.

Revision as of 16:38, 13 March 2015

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, \[\sin x \cos x = \frac{1}{10}.\qquad (*)\]

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*}

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 • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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