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Difference between revisions of "1999 AIME Problems/Problem 11"

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Let <math>s = \sum_{k=1}^{35}\sin 5k  = \sin 5 + \sin 10 + \ldots + \sin 175</math>. We could try to manipulate this sum by wrapping the terms around (since the first half is equal to the second half), but it quickly becomes apparent that this way is difficult to pull off. Instead, we look to [[telescope]] the sum. Using the [[trigonometric identity|identity]] <math>\sin a \sin b = \frac 12(\cos (a-b) - \cos (a+b))</math>, we can rewrite <math>s</math> as  
 
Let <math>s = \sum_{k=1}^{35}\sin 5k  = \sin 5 + \sin 10 + \ldots + \sin 175</math>. We could try to manipulate this sum by wrapping the terms around (since the first half is equal to the second half), but it quickly becomes apparent that this way is difficult to pull off. Instead, we look to [[telescope]] the sum. Using the [[trigonometric identity|identity]] <math>\sin a \sin b = \frac 12(\cos (a-b) - \cos (a+b))</math>, we can rewrite <math>s</math> as  
  
−
<cmath>s \cdot \sin 5 = \sum_{k=1}^{35} \sin 5k \sin 5 = \sum_{k=1}^{35} \frac{1}{2}(\cos (5k - 5)- \cos (5k + 5))</cmath>
+
<cmath>
−
<cmath>s = \frac{0.5(\cos 0 - \cos 10 + \cos 5 - \cos 15 + \cos 10 - \cos 20 + \ldots - \cos 170 + \cos 165 - \cos 175+ \cos 170 - \cos 180)}{\sin 5}</cmath>
+
\begin{align*}
 +
s \cdot \sin 5 = \sum_{k=1}^{35} \sin 5k \sin 5 &= \sum_{k=1}^{35} \frac{1}{2}(\cos (5k - 5)- \cos (5k + 5))\\
 +
&= 0.5(\cos 0 - \cos 10 + \cos 5 - \cos 15 + \cos 10 \ldots + \cos 165 - \cos 175+ \cos 170 - \cos 180)
 +
\end{align*}</cmath>
  
−
This telescopes to <math>s = \frac{\cos 0 + \cos 5 - \cos 175 - \cos 180}{2 \sin 5} = \frac{1 + \cos 5}{\sin 5}</math>. Manipulating this to use the identity <math>\tan x = \frac{1 - \cos 2x}{\sin 2x}</math>, we get <math>s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2}</math>, and our answer is <math>\boxed{177}</math>.
+
This telescopes to <cmath>s = \frac{\cos 0 + \cos 5 - \cos 175 - \cos 180}{2 \sin 5} = \frac{1 + \cos 5}{\sin 5}.</cmath> Manipulating this to use the identity <math>\tan x = \frac{1 - \cos 2x}{\sin 2x}</math>, we get <cmath>s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},</cmath> and our answer is <math>\boxed{177}</math>.
  
 
==Alternate Solution==
 
==Alternate Solution==
  
−
We note that <math>\sin x = \mbox{Im } e^{ix}</math>. We thus have that
+
We note that <math>\sin x = \mbox{Im } e^{ix}\text{*}</math>. We thus have that
−
<math>\sum_{k = 1}^{35} \sin 5k</math>
+
<cmath>\begin{align*}
−
<math>= \sum_{k = 1}^{35} \mbox{Im } e^{5ki} </math>
+
\sum_{k = 1}^{35} \sin 5k &= \sum_{k = 1}^{35} \mbox{Im } e^{5ki}\\
−
<math>= \mbox{Im } \sum_{k = 1}^{35} e^{5ki} </math>
+
&= \mbox{Im } \sum_{k = 1}^{35} e^{5ki}\\
−
<math>= \mbox{Im } \frac{e^{5i}(1 - e^{180i})}{1 - e^{5i}} </math>
+
&= \mbox{Im } \frac{e^{5i}(1 - e^{180i})}{1 - e^{5i}}\\
−
<math>= \mbox{Im } \frac{2\cos5 + 2i \sin 5}{(1 - \cos 5) - i \sin 5} </math>
+
&= \mbox{Im } \frac{2\cos5 + 2i \sin 5}{(1 - \cos 5) - i \sin 5}\\
−
<math>= \mbox{Im } \frac{(2 \cos 5 + 2i \sin 5)((1 - \cos 5) + i \sin 5)}{(1 - \cos 5)^2 + \sin^2 5} </math>
+
&= \mbox{Im } \frac{(2 \cos 5 + 2i \sin 5)[(1 - \cos 5) + i \sin 5]}{(1 - \cos 5)^2 + \sin^2 5}\\
−
<math>= \frac{2 \sin 5}{2 - 2 \cos 5} = \frac{\sin 5}{1 - \cos 5}</math>
+
&= \frac{2 \sin 5}{2 - 2 \cos 5}\\
−
<math>= \frac{\sin 175}{1 + \cos 175} = \tan \frac{175}{2}.</math>  
+
&= \frac{\sin 5}{1 - \cos 5}\\
 +
&= \frac{\sin 175}{1 + \cos 175} \\
 +
&= \tan \frac{175}{2}.\\
 +
\end{align*}</cmath>
 +
The desired answer is thus <math>175 + 2 = \boxed{177}</math>.
 +
 
 +
*Only if <math>x</math> is in radians, which it is not. However, the solution is still viable, so keep reading.
 +
==Solution 3==
 +
Let <math>x=e^{\frac{i\pi}{36}}</math>. By Euler's Formula, <math>\sin{5k^\circ}=\frac{x^k-\frac{1}{x^{k}}}{2i}</math>.
 +
 
 +
The sum we want is thus <math>\frac{x-\frac{1}{x}}{2i}+\frac{x^2-\frac{1}{x^{2}}}{2i}+\cdots+\frac{x^{35}-\frac{1}{x^{35}}}{2i}</math>
 +
 
 +
We factor the <math>\frac{1}{2i}</math> and split into two geometric series to get <math>\frac{1}{2i}\left(\frac{-\frac{1}{x^{35}}(x^{35}-1)}{x-1}+\frac{x(x^{35}-1)}{x-1}\right)</math>
  
 +
However, we note that <math>x^{36}=-1</math>, so <math>-\frac{1}{x^{35}}=x</math>, so our two geometric series are actually the same. We combine the terms and simplify to get <math>\frac{1}{i}\left(\frac{x^{36}-x}{x-1}\right)</math>
  
−
The desired answer is thus <math>175 + 2 = \boxed{177}</math>.
+
Apply Euler's identity and simplify again to get <math>\frac{1}{i}\left(\frac{-x-1}{x-1}\right)</math>
 +
 
 +
Now, we need to figure out how to express this as the tangent of something. We note that <math>\tan(5k^\circ)=\frac{\sin(5k^\circ)}{\cos(5k^\circ)}=\frac{\frac{x^k-\frac{1}{x^k}}{2i}}{\frac{x^k+\frac{1}{x^k}}{2}}=\frac{1}{i}\frac{x^{2k}-1}{x^{2k}+1}</math>.
  
−
Comment: I think it should be <math> \mbox{Im } \sum_{k = 1}^{35} e^{5ki} = \mbox{Im } \frac{e^{5i}(1 - e^{175i})}{1 - e^{5i}} </math>
+
So, we set the two equal to each other to solve for <math>k</math>. Cross multiplying gets <math>(-x-1)(x^{2k}+1)=(x-1)(x^{2k}-1)</math>. Expanding yields <math>-x^{2k+1}-x-x^{2k}-1=x^{2k+1}-x-x^{2k}+1</math>. Simplifying yields <math>x^{2k+1}=-1</math>. Since <math>2k+1=36</math> is the smallest solution, we have <math>k=\frac{35}{2}</math>, and the argument of tangent is <math>5k=\frac{175}{2}</math>. The requested sum is <math>175+2=\boxed{177}</math>.
 +
==Solution 4==
 +
This solution is inspired by [[1997 AIME Problems/Problem 11]].
 +
First of all, <math>\sum_{k=1}^{35}\sin 5k=\sum_{k=0}^{35}\sin 5k</math> because the difference is <math>\sin 0 = 0</math>. Now consider the sum <math>\sum_{k=0}^{35}\cos 5k</math>. Since <math>\cos 5=-\cos 175</math>, <math>\cos 10=-\cos 170</math> and so on, this simplifies to <math>\cos 0=1</math>.
 +
Now, consider the ratio
 +
<cmath>\dfrac{\sum\limits_{n=0}^{35}\sin 5k}{\sum\limits_{n=0}^{35}\cos 5k.}</cmath>
 +
Using the sum-to-product identities, the numerator can be written as
 +
<cmath>\sin \frac{175}{2} (\cos \frac{175}{2} + \cos \frac {165}{2}+\cdots + \cos \frac{5}{2}.)</cmath>
 +
Similarly (via sum-to-product identities), the denominator is equivalent to
 +
<cmath>\cos \frac{175}{2} (\cos \frac{175}{2} + \cos \frac {165}{2}+\cdots + \cos \frac{5}{2}.)</cmath>
 +
If we substitute these equivalent expressions into the original ratio, many parts cancel. We are left with
 +
<cmath>\frac{\sin \frac{175}{2}}{\cos \frac{175}{2}} = \tan{\frac{175}{2}}.</cmath>
 +
Aha! Remembering that the original denominator <math>\sum_{k=0}^{35}\cos 5k</math> was equal to 1, we realize that the numerator, <math>\sum_{k=0}^{35}\sin 5k=\tan{\frac{175}{2}}</math>. Since <math>\sum_{k=0}^{35}\sin 5k=\sum_{k=1}^{35}\sin 5k</math>, we get <math>\tan \frac{175}{2}</math>, or <math>m+n=\boxed{177}</math>, as our final answer. <math>\blacksquare</math>
 +
~ewei12
  
 
== See also ==
 
== See also ==

Latest revision as of 12:39, 29 June 2024

Problem

Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}$ and $n_{}$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$

Solution

Let $s = \sum_{k=1}^{35}\sin 5k = \sin 5 + \sin 10 + \ldots + \sin 175$. We could try to manipulate this sum by wrapping the terms around (since the first half is equal to the second half), but it quickly becomes apparent that this way is difficult to pull off. Instead, we look to telescope the sum. Using the identity $\sin a \sin b = \frac 12(\cos (a-b) - \cos (a+b))$, we can rewrite $s$ as

\begin{align*} s \cdot \sin 5 = \sum_{k=1}^{35} \sin 5k \sin 5 &= \sum_{k=1}^{35} \frac{1}{2}(\cos (5k - 5)- \cos (5k + 5))\\ &= 0.5(\cos 0 - \cos 10 + \cos 5 - \cos 15 + \cos 10 \ldots + \cos 165 - \cos 175+ \cos 170 - \cos 180) \end{align*}

This telescopes to \[s = \frac{\cos 0 + \cos 5 - \cos 175 - \cos 180}{2 \sin 5} = \frac{1 + \cos 5}{\sin 5}.\] Manipulating this to use the identity $\tan x = \frac{1 - \cos 2x}{\sin 2x}$, we get \[s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},\] and our answer is $\boxed{177}$.

Alternate Solution

We note that $\sin x = \mbox{Im } e^{ix}\text{*}$. We thus have that \begin{align*} \sum_{k = 1}^{35} \sin 5k &= \sum_{k = 1}^{35} \mbox{Im } e^{5ki}\\ &= \mbox{Im } \sum_{k = 1}^{35} e^{5ki}\\ &= \mbox{Im } \frac{e^{5i}(1 - e^{180i})}{1 - e^{5i}}\\ &= \mbox{Im } \frac{2\cos5 + 2i \sin 5}{(1 - \cos 5) - i \sin 5}\\ &= \mbox{Im } \frac{(2 \cos 5 + 2i \sin 5)[(1 - \cos 5) + i \sin 5]}{(1 - \cos 5)^2 + \sin^2 5}\\ &= \frac{2 \sin 5}{2 - 2 \cos 5}\\ &= \frac{\sin 5}{1 - \cos 5}\\ &= \frac{\sin 175}{1 + \cos 175} \\ &= \tan \frac{175}{2}.\\ \end{align*} The desired answer is thus $175 + 2 = \boxed{177}$.

  • Only if $x$ is in radians, which it is not. However, the solution is still viable, so keep reading.

Solution 3

Let $x=e^{\frac{i\pi}{36}}$. By Euler's Formula, $\sin{5k^\circ}=\frac{x^k-\frac{1}{x^{k}}}{2i}$.

The sum we want is thus $\frac{x-\frac{1}{x}}{2i}+\frac{x^2-\frac{1}{x^{2}}}{2i}+\cdots+\frac{x^{35}-\frac{1}{x^{35}}}{2i}$

We factor the $\frac{1}{2i}$ and split into two geometric series to get $\frac{1}{2i}\left(\frac{-\frac{1}{x^{35}}(x^{35}-1)}{x-1}+\frac{x(x^{35}-1)}{x-1}\right)$

However, we note that $x^{36}=-1$, so $-\frac{1}{x^{35}}=x$, so our two geometric series are actually the same. We combine the terms and simplify to get $\frac{1}{i}\left(\frac{x^{36}-x}{x-1}\right)$

Apply Euler's identity and simplify again to get $\frac{1}{i}\left(\frac{-x-1}{x-1}\right)$

Now, we need to figure out how to express this as the tangent of something. We note that $\tan(5k^\circ)=\frac{\sin(5k^\circ)}{\cos(5k^\circ)}=\frac{\frac{x^k-\frac{1}{x^k}}{2i}}{\frac{x^k+\frac{1}{x^k}}{2}}=\frac{1}{i}\frac{x^{2k}-1}{x^{2k}+1}$.

So, we set the two equal to each other to solve for $k$. Cross multiplying gets $(-x-1)(x^{2k}+1)=(x-1)(x^{2k}-1)$. Expanding yields $-x^{2k+1}-x-x^{2k}-1=x^{2k+1}-x-x^{2k}+1$. Simplifying yields $x^{2k+1}=-1$. Since $2k+1=36$ is the smallest solution, we have $k=\frac{35}{2}$, and the argument of tangent is $5k=\frac{175}{2}$. The requested sum is $175+2=\boxed{177}$.

Solution 4

This solution is inspired by 1997 AIME Problems/Problem 11. First of all, $\sum_{k=1}^{35}\sin 5k=\sum_{k=0}^{35}\sin 5k$ because the difference is $\sin 0 = 0$. Now consider the sum $\sum_{k=0}^{35}\cos 5k$. Since $\cos 5=-\cos 175$, $\cos 10=-\cos 170$ and so on, this simplifies to $\cos 0=1$. Now, consider the ratio \[\dfrac{\sum\limits_{n=0}^{35}\sin 5k}{\sum\limits_{n=0}^{35}\cos 5k.}\] Using the sum-to-product identities, the numerator can be written as \[\sin \frac{175}{2} (\cos \frac{175}{2} + \cos \frac {165}{2}+\cdots + \cos \frac{5}{2}.)\] Similarly (via sum-to-product identities), the denominator is equivalent to \[\cos \frac{175}{2} (\cos \frac{175}{2} + \cos \frac {165}{2}+\cdots + \cos \frac{5}{2}.)\] If we substitute these equivalent expressions into the original ratio, many parts cancel. We are left with \[\frac{\sin \frac{175}{2}}{\cos \frac{175}{2}} = \tan{\frac{175}{2}}.\] Aha! Remembering that the original denominator $\sum_{k=0}^{35}\cos 5k$ was equal to 1, we realize that the numerator, $\sum_{k=0}^{35}\sin 5k=\tan{\frac{175}{2}}$. Since $\sum_{k=0}^{35}\sin 5k=\sum_{k=1}^{35}\sin 5k$, we get $\tan \frac{175}{2}$, or $m+n=\boxed{177}$, as our final answer. $\blacksquare$ ~ewei12

See also

1999 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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