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2015 AIME II Problems/Problem 13 - Revision history
2024-03-28T11:00:49Z
Revision history for this page on the wiki
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https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_13&diff=210758&oldid=prev
Mathboy282: /* Clearer Explanation for the End */
2024-01-15T01:29:45Z
<p><span dir="auto"><span class="autocomment">Clearer Explanation for the End</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 01:29, 15 January 2024</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l9" >Line 9:</td>
<td colspan="2" class="diff-lineno">Line 9:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Since <math>2\sin 1</math> is positive, it does not affect the sign of <math>a_n</math>.  Let <math>b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)</math>.  Now since <math>\cos(0) + \cos(1) = 2\cos\left(\cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math> and <math>\cos(n) + \cos(n + 1) = 2\cos\left(n + \cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math>, <math>b_n</math> is negative if and only if <math>\cos\left(\cfrac{1}{2}\right) < \cos\left(n + \cfrac{1}{2}\right)</math>, or when <math>n \in [2k\pi - 1, 2k\pi]</math>.  Since <math>\pi</math> is irrational, there is always only one integer in the range, so there are values of <math>n</math> such that <math>a_n < 0</math> at <math>2\pi, 4\pi, \cdots</math>.  Then the hundredth such value will be when <math>k = 100</math> and <math>n = \lfloor 200\pi \rfloor = \lfloor 628.318 \rfloor = \boxed{628}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Since <math>2\sin 1</math> is positive, it does not affect the sign of <math>a_n</math>.  Let <math>b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)</math>.  Now since <math>\cos(0) + \cos(1) = 2\cos\left(\cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math> and <math>\cos(n) + \cos(n + 1) = 2\cos\left(n + \cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math>, <math>b_n</math> is negative if and only if <math>\cos\left(\cfrac{1}{2}\right) < \cos\left(n + \cfrac{1}{2}\right)</math>, or when <math>n \in [2k\pi - 1, 2k\pi]</math>.  Since <math>\pi</math> is irrational, there is always only one integer in the range, so there are values of <math>n</math> such that <math>a_n < 0</math> at <math>2\pi, 4\pi, \cdots</math>.  Then the hundredth such value will be when <math>k = 100</math> and <math>n = \lfloor 200\pi \rfloor = \lfloor 628.318 \rfloor = \boxed{628}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Clearer Explanation for the End===</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Clearer Explanation for the End===</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Notice that we want <math>\cos(n+1/2)>cos(1/2).</math></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Notice that we want <math>\cos(n+1/2)><ins class="diffchange diffchange-inline">\</ins>cos(1/2).</math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that the only interval in which this is true, is where <math>-1/2 < n + 1/2 < 1/2.</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that the only interval in which this is true, is where <math>-1/2 < n + 1/2 < 1/2.</math></div></td></tr>
</table>
Mathboy282
https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_13&diff=210757&oldid=prev
Mathboy282: /* Solution 1 */
2024-01-15T01:29:27Z
<p><span dir="auto"><span class="autocomment">Solution 1</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 01:29, 15 January 2024</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l8" >Line 8:</td>
<td colspan="2" class="diff-lineno">Line 8:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath>a_n = \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{2\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{2\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath>a_n = \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{2\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{2\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].</cmath></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Since <math>2\sin 1</math> is positive, it does not affect the sign of <math>a_n</math>.  Let <math>b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)</math>.  Now since <math>\cos(0) + \cos(1) = 2\cos\left(\cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math> and <math>\cos(n) + \cos(n + 1) = 2\cos\left(n + \cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math>, <math>b_n</math> is negative if and only if <math>\cos\left(\cfrac{1}{2}\right) < \cos\left(n + \cfrac{1}{2}\right)</math>, or when <math>n \in [2k\pi - 1, 2k\pi]</math>.  Since <math>\pi</math> is irrational, there is always only one integer in the range, so there are values of <math>n</math> such that <math>a_n < 0</math> at <math>2\pi, 4\pi, \cdots</math>.  Then the hundredth such value will be when <math>k = 100</math> and <math>n = \lfloor 200\pi \rfloor = \lfloor 628.318 \rfloor = \boxed{628}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Since <math>2\sin 1</math> is positive, it does not affect the sign of <math>a_n</math>.  Let <math>b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)</math>.  Now since <math>\cos(0) + \cos(1) = 2\cos\left(\cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math> and <math>\cos(n) + \cos(n + 1) = 2\cos\left(n + \cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math>, <math>b_n</math> is negative if and only if <math>\cos\left(\cfrac{1}{2}\right) < \cos\left(n + \cfrac{1}{2}\right)</math>, or when <math>n \in [2k\pi - 1, 2k\pi]</math>.  Since <math>\pi</math> is irrational, there is always only one integer in the range, so there are values of <math>n</math> such that <math>a_n < 0</math> at <math>2\pi, 4\pi, \cdots</math>.  Then the hundredth such value will be when <math>k = 100</math> and <math>n = \lfloor 200\pi \rfloor = \lfloor 628.318 \rfloor = \boxed{628}</math>.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">===Clearer Explanation for the End===</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Notice that we want <math>\cos(n+1/2)>cos(1/2).</math></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Notice that the only interval in which this is true, is where <math>-1/2 < n + 1/2 < 1/2.</math></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Thus, we need <math>-1<n<0.</math> So we have <math>-1+2\pi k <n <0+2\pi k.</math></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">But we want the 100th solution. Which means <math>n</math> is in between <math>200\pi - 1 < n < 200 \pi.</math> So n = 628.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">~mathboy282</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 2==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 2==</div></td></tr>
</table>
Mathboy282
https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_13&diff=188463&oldid=prev
Daijobu at 01:01, 2 February 2023
2023-02-02T01:01:11Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 01:01, 2 February 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l27" >Line 27:</td>
<td colspan="2" class="diff-lineno">Line 27:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>So the argument of <math>z^n</math> must be in the bound <math>-1<\theta<0</math> by doubling, namely the last <math>z^n</math> negative before another rotation. Since there is always one <math>z^n</math> in this category per rotation because <math>\pi</math> is irrational, <math>n_{100}\equiv z^{628}</math> and the answer is <math>\boxed{628}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>So the argument of <math>z^n</math> must be in the bound <math>-1<\theta<0</math> by doubling, namely the last <math>z^n</math> negative before another rotation. Since there is always one <math>z^n</math> in this category per rotation because <math>\pi</math> is irrational, <math>n_{100}\equiv z^{628}</math> and the answer is <math>\boxed{628}</math>.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">==Video Solution==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">https://youtu.be/b1-cUUPjYNk</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">~MathProblemSolvingSkills</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==See also==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==See also==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{AIME box|year=2015|n=II|num-b=12|num-a=14}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{AIME box|year=2015|n=II|num-b=12|num-a=14}}</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td></tr>
</table>
Daijobu
https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_13&diff=110046&oldid=prev
Rzlng: /* Solution 2 */
2019-09-28T23:09:34Z
<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
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<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:09, 28 September 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l12" >Line 12:</td>
<td colspan="2" class="diff-lineno">Line 12:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math>\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(<del class="diffchange diffchange-inline">k </del>+ \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( n + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),</math> by sum to product. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math>\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(<ins class="diffchange diffchange-inline">n </ins>+ \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( n + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),</math> by sum to product. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 3==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 3==</div></td></tr>
</table>
Rzlng
https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_13&diff=110045&oldid=prev
Rzlng: /* Solution 2 */
2019-09-28T23:09:20Z
<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:09, 28 September 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l12" >Line 12:</td>
<td colspan="2" class="diff-lineno">Line 12:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math>\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(<del class="diffchange diffchange-inline">n </del>+ \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( n + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),</math> by sum to product. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math>\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(<ins class="diffchange diffchange-inline">k </ins>+ \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( n + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),</math> by sum to product. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 3==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 3==</div></td></tr>
</table>
Rzlng
https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_13&diff=110044&oldid=prev
Rzlng: /* Solution 2 */
2019-09-28T23:09:11Z
<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:09, 28 September 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l12" >Line 12:</td>
<td colspan="2" class="diff-lineno">Line 12:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math>\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(<del class="diffchange diffchange-inline">k </del>+ \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( n + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),</math> by sum to product. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math>\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(<ins class="diffchange diffchange-inline">n </ins>+ \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( n + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),</math> by sum to product. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 3==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 3==</div></td></tr>
</table>
Rzlng
https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_13&diff=110043&oldid=prev
Rzlng: /* Solution 2 */
2019-09-28T23:08:56Z
<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:08, 28 September 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l12" >Line 12:</td>
<td colspan="2" class="diff-lineno">Line 12:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math>\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(k + \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( <del class="diffchange diffchange-inline">k </del>+ \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),</math> by sum to product. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math>\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(k + \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( <ins class="diffchange diffchange-inline">n </ins>+ \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),</math> by sum to product. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 3==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 3==</div></td></tr>
</table>
Rzlng
https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_13&diff=110042&oldid=prev
Rzlng: /* Solution 2 */
2019-09-28T23:03:55Z
<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:03, 28 September 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l12" >Line 12:</td>
<td colspan="2" class="diff-lineno">Line 12:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math><del class="diffchange diffchange-inline">\sin n + \sin 1 < \sin (n+1) \implies </del>\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(k + \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( k + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),</math> by sum to product. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math>\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(k + \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( k + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),</math> by sum to product. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 3==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 3==</div></td></tr>
</table>
Rzlng
https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_13&diff=110041&oldid=prev
Rzlng: /* Solution 2 */
2019-09-28T23:03:20Z
<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:03, 28 September 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l12" >Line 12:</td>
<td colspan="2" class="diff-lineno">Line 12:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice that <math>a_n</math> is the imaginary part of <math>\sum_{k=1}^n e^{ik}</math>, by Euler's formula. Using the geometric series formula, we find that this sum is equal to <cmath>\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}</cmath> We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have <cmath> \frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}</cmath> We only need to look at the imaginary part, which is <cmath>\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}</cmath></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math>\sin n + \sin 1 < \sin (n+1) \implies \sin (n+1) - \sin n > \sin 1</math>. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Since <math>\cos 1 < 1</math>, <math>2-2 \cos 1 > 0</math>, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have <math>\sin n + \sin 1 < \sin (n+1) \implies \sin (n+1) - \sin n > \sin 1 <ins class="diffchange diffchange-inline">\implies 2 \cos \left(k + \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( k + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),</ins></math> <ins class="diffchange diffchange-inline">by sum to product</ins>. This only holds when <math>n</math> is between <math>2\pi k - 1</math> and <math>2\pi k</math> for integer <math>k</math> [continuity proof here], and since this has exactly one integer solution for every such interval, the <math>100</math>th such <math>n</math> is <math>\lfloor 200\pi \rfloor = \boxed{628}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 3==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 3==</div></td></tr>
</table>
Rzlng
https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_13&diff=99961&oldid=prev
E power pi times i: Product to sum identity was written wrong.
2019-01-02T06:34:42Z
<p>Product to sum identity was written wrong.</p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 06:34, 2 January 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l6" >Line 6:</td>
<td colspan="2" class="diff-lineno">Line 6:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>If <math>n = 1</math>, <math>a_n = \sin(1) > 0</math>.  Then if <math>n</math> satisfies <math>a_n < 0</math>, <math>n \ge 2</math>, and</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>If <math>n = 1</math>, <math>a_n = \sin(1) > 0</math>.  Then if <math>n</math> satisfies <math>a_n < 0</math>, <math>n \ge 2</math>, and</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><cmath>a_n = \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].</cmath></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><cmath>a_n = \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{<ins class="diffchange diffchange-inline">2</ins>\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{<ins class="diffchange diffchange-inline">2</ins>\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].</cmath></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Since <math>\sin 1</math> is positive, it does not affect the sign of <math>a_n</math>.  Let <math>b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)</math>.  Now since <math>\cos(0) + \cos(1) = 2\cos\left(\cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math> and <math>\cos(n) + \cos(n + 1) = 2\cos\left(n + \cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math>, <math>b_n</math> is negative if and only if <math>\cos\left(\cfrac{1}{2}\right) < \cos\left(n + \cfrac{1}{2}\right)</math>, or when <math>n \in [2k\pi - 1, 2k\pi]</math>.  Since <math>\pi</math> is irrational, there is always only one integer in the range, so there are values of <math>n</math> such that <math>a_n < 0</math> at <math>2\pi, 4\pi, \cdots</math>.  Then the hundredth such value will be when <math>k = 100</math> and <math>n = \lfloor 200\pi \rfloor = \lfloor 628.318 \rfloor = \boxed{628}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Since <math><ins class="diffchange diffchange-inline">2</ins>\sin 1</math> is positive, it does not affect the sign of <math>a_n</math>.  Let <math>b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)</math>.  Now since <math>\cos(0) + \cos(1) = 2\cos\left(\cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math> and <math>\cos(n) + \cos(n + 1) = 2\cos\left(n + \cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)</math>, <math>b_n</math> is negative if and only if <math>\cos\left(\cfrac{1}{2}\right) < \cos\left(n + \cfrac{1}{2}\right)</math>, or when <math>n \in [2k\pi - 1, 2k\pi]</math>.  Since <math>\pi</math> is irrational, there is always only one integer in the range, so there are values of <math>n</math> such that <math>a_n < 0</math> at <math>2\pi, 4\pi, \cdots</math>.  Then the hundredth such value will be when <math>k = 100</math> and <math>n = \lfloor 200\pi \rfloor = \lfloor 628.318 \rfloor = \boxed{628}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 2==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 2==</div></td></tr>
</table>
E power pi times i