Difference between revisions of "Talk:2013 AIME I Problems/Problem 14"

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For solution 1 where they talk about <math>P \sin \theta + Q \cos \theta = \cos \theta - \frac{1}{2} P</math>, I got <cmath> \cos \theta - \frac{1}{4} \cos \theta + \frac{1}{8} \sin 2 \theta - \frac{1}{16} \cos 3 \theta + \frac{1}{32} \sin 4 \theta - \dots </cmath> instead (the same as in solution 2), which is not what the solution 1 got. If the writer of solution 1 could resolve this, that would be great.
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For solution 1 where they talk about <math>P \sin \theta + Q \cos \theta = \cos \theta - \dfrac{1}{4} \cos \theta - \dfrac{1}{8} \sin{2 \theta} - \dfrac{1}{16} \cos{3 \theta} + \cdots</math>, I got <cmath> \cos \theta - \frac{1}{4} \cos \theta + \frac{1}{8} \sin 2 \theta - \frac{1}{16} \cos 3 \theta + \frac{1}{32} \sin 4 \theta - \dots </cmath> instead (the same as in solution 2), which is not what the solution 1 got. If the writer of solution 1 could resolve this, that would be great.
  
 
~MeepMurp5
 
~MeepMurp5

Revision as of 13:12, 27 December 2021

For solution 1 where they talk about $P \sin \theta + Q \cos \theta = \cos \theta - \dfrac{1}{4} \cos \theta - \dfrac{1}{8} \sin{2 \theta} - \dfrac{1}{16} \cos{3 \theta} + \cdots$, I got \[\cos \theta - \frac{1}{4} \cos \theta + \frac{1}{8} \sin 2 \theta - \frac{1}{16} \cos 3 \theta + \frac{1}{32} \sin 4 \theta - \dots\] instead (the same as in solution 2), which is not what the solution 1 got. If the writer of solution 1 could resolve this, that would be great.

~MeepMurp5