2003 AIME I Problems/Problem 4
Revision as of 18:58, 4 July 2013 by Nathan wailes (talk | contribs)
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.
\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, , 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.