Difference between revisions of "2003 AIME I Problems/Problem 4"
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== Problem == | == Problem == | ||
| − | 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 - | + | 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 - 10), </math> find <math> n. </math> |
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== 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, <cmath> \sin x \cos x = \frac{1}{10}.\qquad (*)</cmath> | 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> | ||
Revision as of 13:23, 26 February 2017
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.