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 - 1), </math> find <math> n. </math> |
− | == Solution == | + | == Solution 1 == |
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> | ||
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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>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | Examining the first equation, we simplify as the following: | ||
+ | <cmath>\log_{10} \sin x \cos x = -1</cmath> | ||
+ | <cmath>\implies \sin x \cos x = \frac{1}{10}</cmath> | ||
+ | |||
+ | With this in mind, examining the second equation, we may simplify as the following (utilizing logarithm properties): | ||
+ | <cmath>\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - \log_{10} 10)</cmath> | ||
+ | <cmath>\implies \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} \frac{n}{10})</cmath> | ||
+ | <cmath>\implies \log_{10} (\sin x + \cos x) = \log_{10} \sqrt{\frac{n}{10}}</cmath> | ||
+ | |||
+ | From here, we may divide both sides by <math>\log_{10} (\sin x + \cos x)</math> and then proceed with the change-of-base logarithm property: | ||
+ | <cmath>1 = \frac{\log_{10} \sqrt{\frac{n}{10}}}{\log_{10} (\sin x + \cos x)}</cmath> | ||
+ | <cmath>\implies 1 = \log_{\sin x + \cos x} \sqrt{\frac{n}{10}}</cmath> | ||
+ | |||
+ | Thus, exponentiating both sides results in <math>\sin x + \cos x = \sqrt{\frac{n}{10}}</math>. Squaring both sides gives us | ||
+ | <cmath>\sin^2 x + 2\sin x \cos x + \cos^2 x = \frac{n}{10}</cmath> | ||
+ | |||
+ | Via the Pythagorean Identity, <math>\sin^2 x + \cos^2 x = 1</math> and <math>2\sin x \cos x</math> is simply <math>\frac{1}{5}</math>, via substitution. Thus, substituting these results into the current equation: | ||
+ | <cmath>1 + \frac{1}{5} = \frac{n}{10}</cmath> | ||
+ | <cmath>\implies \frac{6}{5} = \frac{n}{10}</cmath> | ||
+ | |||
+ | Using simple cross-multiplication techniques, we have <math>5n = 60</math>, and thus <math>\boxed{n = 012}</math>. | ||
+ | ~ nikenissan | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | By the first equation, we get that <math>\sin(x)*\cos(x)=10^{-1}</math>. We can let <math>\sin(x)=a</math>, <math>\cos(x)=b</math>. Thus <math>ab=\frac{1}{10}</math>. By the identity <math>\sin^2x+\cos^2x=1</math>, we get that <math>a^2+b^2=1</math>. Solving this, we get <math>a+b=\sqrt{\frac{12}{10}}</math>. So we have | ||
+ | |||
+ | <cmath>\log\left(\sqrt{\frac{12}{10}}\right)=\frac12(\log(n)-1)</cmath> | ||
+ | <cmath>2\log\left(\sqrt{\frac{12}{10}}\right)=\log(n)-1</cmath> | ||
+ | <cmath>\log\left(\frac{12}{10}\right)+1=\log(n)</cmath> | ||
+ | <cmath>\log\left(\frac{12}{10}\right)+\log(10)=\log(n)</cmath> | ||
+ | <cmath>\log\left(\frac{12}{10}\times 10\right)=\log(12)=\log(n)</cmath> | ||
+ | |||
+ | From here it is obvious that <math>\boxed{n=012}</math>. | ||
+ | |||
+ | |||
+ | ~yofro | ||
== See also == | == See also == |
Latest revision as of 12:32, 10 June 2024
Problem
Given that and that
find
Solution 1
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
.
.
Solution 2
Examining the first equation, we simplify as the following:
With this in mind, examining the second equation, we may simplify as the following (utilizing logarithm properties):
From here, we may divide both sides by and then proceed with the change-of-base logarithm property:
Thus, exponentiating both sides results in . Squaring both sides gives us
Via the Pythagorean Identity, and
is simply
, via substitution. Thus, substituting these results into the current equation:
Using simple cross-multiplication techniques, we have , and thus
.
~ nikenissan
Solution 3
By the first equation, we get that . We can let
,
. Thus
. By the identity
, we get that
. Solving this, we get
. So we have
From here it is obvious that .
~yofro
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.