Difference between revisions of "2003 AIME I Problems/Problem 4"
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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, < | + | 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. | ||
− | < | + | <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) \\ | ||
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\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*} | ||
− | </ | + | </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 and that find
Solution
Using the properties of logarithms, we can simplify the first equation to . Therefore,
Now, manipulate the second equation.
By the Pythagorean identities, , and we can substitute the value for from . .
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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