Difference between revisions of "2015 AIME I Problems/Problem 13"
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Putting this back into our equation and simplifying gives us <math>\sqrt{\frac{1}{2p}}=\frac{1}{2^{45}}\prod_{k=23}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \cos(2k)^\circ</math>. | Putting this back into our equation and simplifying gives us <math>\sqrt{\frac{1}{2p}}=\frac{1}{2^{45}}\prod_{k=23}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \cos(2k)^\circ</math>. | ||
− | Using the fact that <math>\sin\theta=\cos( | + | Using the fact that <math>\sin\theta=\cos(90^{\circ}-\theta)</math> again, our equation simplifies to <math>\sqrt{\frac{1}{2p}}=\frac{\sin90^\circ}{2^{45}}</math>, and since <math>\sin90^\circ=1</math>, it follows that <math>2p = 2^{90}</math>, which implies <math>p=2^{89}</math>. Thus, <math>m+n=2+89=\boxed{091}</math>. |
==See Also== | ==See Also== |
Revision as of 14:02, 19 February 2016
Contents
[hide]Problem
With all angles measured in degrees, the product , where and are integers greater than 1. Find .
Solution
Solution 1
Let . Then from the identity we deduce that (taking absolute values and noticing ) But because is the reciprocal of and because , if we let our product be then because is positive in the first and second quadrants. Now, notice that are the roots of Hence, we can write , and so It is easy to see that and that our answer is .
Solution 2
Let
because of the identity
we want
Thus the answer is
Solution 3
Similar to Solution , so we use and we find that: Now we can cancel the sines of the multiples of : So and we can apply the double-angle formula again: Of course, is missing, so we multiply it to both sides: Now isolate the product of the sines: And the product of the squares of the cosecants as asked for by the problem is the square of the inverse of this number: The answer is therefore .
Solution 4
Let .
Then, .
Since , we can multiply both sides by to get .
Using the double-angle identity , we get .
Note that the right-hand side is equal to , which is equal to , again, from using our double-angle identity.
Putting this back into our equation and simplifying gives us .
Using the fact that again, our equation simplifies to , and since , it follows that , which implies . Thus, .
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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