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

(prosify, box)
(teX cleanup)
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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 ==
−
The first equation, <math>\displaystyle \log_{10} \sin x + \log_{10} \cos x = -1 </math>, can be combined under the properties of [[logarithm]]s to <math> \displaystyle \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, <math> \sin x \cos x = \frac{1}{10}\ (*)</math>.
  
−
Now, manipulate the second equation, <math> \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1) </math>.
+
Now, manipulate the second equation.
 +
<center><math>\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*}
 +
</math></center>
  
−
:<math> \log_{10} (\sin x + \cos x) = \frac{1}{2}(\log_{10} n - \log_{10} 10) </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>.
−
 
 
−
:<math> \log_{10} (\sin x + \cos x) = \frac{1}{2}(\log_{10} \frac{n}{10}) </math>
 
−
 
 
−
:<math> \log_{10} (\sin x + \cos x) = \left(\log_{10} \sqrt{\frac{n}{10}}\right) </math>
 
−
 
 
−
:<math> \sin x + \cos x = \sqrt{\frac{n}{10}} </math>
 
−
 
 
−
:<math> (\sin x + \cos x)^{2} = \left(\sqrt{\frac{n}{10}}\right)^2 </math>
 
−
 
 
−
:<math> \sin^2 x + \cos^2 x +2 \sin x \cos x= \frac{n}{10} </math>
 
−
 
 
−
<math>\displaystyle \sin ^2 x + \cos ^2 x = 1</math>, and we can substitute the value for <math>\displaystyle \sin x \cos x</math> from above.
 
−
 
 
−
:<math> 1 + 2(\frac{1}{10}) = \frac{n}{10} </math>
 
−
 
 
−
:<math> \frac{12}{10} = \frac{n}{10} </math>
 
−
 
 
−
Thus, the solution is <math> n = 012 </math>.
 
  
 
== See also ==
 
== See also ==
−
* [[Logarithm]]
 
−
* [[Trigonometry]]
 
 
{{AIME box|year=2003|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2003|n=I|num-b=3|num-a=5}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Trigonometry Problems]]
 
[[Category:Intermediate Trigonometry Problems]]

Revision as of 15:04, 10 June 2008

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 $(*)$. $1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} \Longrightarrow n = \boxed{012}$.

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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