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 - \ | + | 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> P \sin \theta + Q \cos \theta = \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 | ||
Latest revision as of 12:19, 27 December 2021
For solution 1 where they talk about 
, I got ![\[P \sin \theta + Q \cos \theta = \cos \theta - \frac{1}{4} \cos \theta + \frac{1}{8} \sin 2 \theta + \frac{1}{16} \cos 3 \theta - \frac{1}{32} \sin 4 \theta + \dots\]](http://latex.artofproblemsolving.com/4/e/9/4e90fe5dd48bce2ffcc8e19a5ffa2b55c33dd5f1.png)
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
