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

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==Solution 4==
 
==Solution 4==
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Let <math>a=\cos(x)</math> and <math>b=\sin(x)</math>. We know that <math>a^2+b^2=1</math> and <math>ab=\frac{1}{10}</math>.
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Let <math>\log{x} = \log_{10}{x}.</math> Through basic log properties, we see that <math>\log{a} + \log{b} = \log{(ab)}.</math> Thus, we see that <math>\log{(\sin{x})} + \log{(\cos{x})} = \log{(\sin{x}\cos{x})} = -1.</math> Simplifying, we get:
 +
 
 +
\begin{align*}
 +
\log{(\sin{x}\cos{x})} &= -1 \\
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\sin{x}\cos{x} &= 10^{-1} = \frac{1}{10}
 +
\end{align*}
 +
 
 +
Next, we can manipulate the second equation to get:
 +
 
 +
\begin{align*}
 +
\log{(\sin{x} + \cos{x})} &= \frac{1}{2}(\log{n}-1) \\
 +
2\log{(\sin{x} + \cos{x})} &= \log{n}-1 \\
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\log{(\sin{x} + \cos{x})^2} + 1 &= \log{n}
 +
\end{align*}
 +
 
 +
Expanding <math>(\sin{x} + \cos{x})^2,</math> we get:
 +
 
 +
\begin{align*}
 +
\log{(\sin^2{x} + \cos^2{x} + 2\sin{x}\cos{x})} + 1 &= \log{n} \\
 +
\log{(1 + 2\sin{x}\cos{x})} + 1 &= \log{n} \\
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\log{(1 + \frac{2}{10})} + \log{10} &= \log{n} \\
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\log{(\frac{12}{10} \cdot 10)} = \log{n} \\
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\log{12} = \log{n}
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\end{align*}
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Finally, we see that <math>n = \boxed{012}.</math>
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~ Cheetahboy93
  
 
== See also ==
 
== See also ==

Latest revision as of 14:30, 22 September 2024

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 1

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, \[\sin x \cos x = \frac{1}{10}.\qquad (*)\]

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*}

By the Pythagorean identities, $\sin ^2 x + \cos ^2 x = 1$, and we can substitute the value for $\sin x \cos x$ from $(*)$. $1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} \Longrightarrow n = \boxed{012}$.

Solution 2

Examining the first equation, we simplify as the following: \[\log_{10} \sin x \cos x = -1\] \[\implies \sin x \cos x = \frac{1}{10}\]

With this in mind, examining the second equation, we may simplify as the following (utilizing logarithm properties): \[\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - \log_{10} 10)\] \[\implies \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} \frac{n}{10})\] \[\implies \log_{10} (\sin x + \cos x) = \log_{10} \sqrt{\frac{n}{10}}\]

From here, we may divide both sides by $\log_{10} (\sin x + \cos x)$ and then proceed with the change-of-base logarithm property: \[1 = \frac{\log_{10} \sqrt{\frac{n}{10}}}{\log_{10} (\sin x + \cos x)}\] \[\implies 1 = \log_{\sin x + \cos x} \sqrt{\frac{n}{10}}\]

Thus, exponentiating both sides results in $\sin x + \cos x = \sqrt{\frac{n}{10}}$. Squaring both sides gives us \[\sin^2 x + 2\sin x \cos x + \cos^2 x = \frac{n}{10}\]

Via the Pythagorean Identity, $\sin^2 x + \cos^2 x = 1$ and $2\sin x \cos x$ is simply $\frac{1}{5}$, via substitution. Thus, substituting these results into the current equation: \[1 + \frac{1}{5} = \frac{n}{10}\] \[\implies \frac{6}{5} = \frac{n}{10}\]

Using simple cross-multiplication techniques, we have $5n = 60$, and thus $\boxed{n = 012}$. ~ nikenissan


Solution 3

By the first equation, we get that $\sin(x)*\cos(x)=10^{-1}$. We can let $\sin(x)=a$, $\cos(x)=b$. Thus $ab=\frac{1}{10}$. By the identity $\sin^2x+\cos^2x=1$, we get that $a^2+b^2=1$. Solving this, we get $a+b=\sqrt{\frac{12}{10}}$. So we have

\[\log\left(\sqrt{\frac{12}{10}}\right)=\frac12(\log(n)-1)\] \[2\log\left(\sqrt{\frac{12}{10}}\right)=\log(n)-1\] \[\log\left(\frac{12}{10}\right)+1=\log(n)\] \[\log\left(\frac{12}{10}\right)+\log(10)=\log(n)\] \[\log\left(\frac{12}{10}\times 10\right)=\log(12)=\log(n)\]

From here it is obvious that $\boxed{n=012}$.


~yofro

Solution 4

Let $\log{x} = \log_{10}{x}.$ Through basic log properties, we see that $\log{a} + \log{b} = \log{(ab)}.$ Thus, we see that $\log{(\sin{x})} + \log{(\cos{x})} = \log{(\sin{x}\cos{x})} = -1.$ Simplifying, we get:

\begin{align*} \log{(\sin{x}\cos{x})} &= -1 \\ \sin{x}\cos{x} &= 10^{-1} = \frac{1}{10} \end{align*}

Next, we can manipulate the second equation to get:

\begin{align*} \log{(\sin{x} + \cos{x})} &= \frac{1}{2}(\log{n}-1) \\ 2\log{(\sin{x} + \cos{x})} &= \log{n}-1 \\ \log{(\sin{x} + \cos{x})^2} + 1 &= \log{n} \end{align*}

Expanding $(\sin{x} + \cos{x})^2,$ we get:

\begin{align*} \log{(\sin^2{x} + \cos^2{x} + 2\sin{x}\cos{x})} + 1 &= \log{n} \\ \log{(1 + 2\sin{x}\cos{x})} + 1 &= \log{n} \\ \log{(1 + \frac{2}{10})} + \log{10} &= \log{n} \\ \log{(\frac{12}{10} \cdot 10)} = \log{n} \\ \log{12} = \log{n} \end{align*}

Finally, we see that $n = \boxed{012}.$

~ Cheetahboy93

See also

2003 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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All AIME Problems and Solutions

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