2015 AIME I Problems/Problem 13
With all angles measured in degrees, the product , where and are integers greater than 1. Find .
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 .
because of the identity
Thus the answer is
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 .
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, .
Once we have the tools of complex polynomials there is no need to use the tactical tricks. Everything is so basic (I think).
Recall that the roots of are , we have Let , and take absolute value of both sides, or, Let be even, then, so, Set and we have , -Mathdummy
Recall that Since it is in csc, we can write in sin and then take reciprocal. We can group them by threes, . Thus So we take reciprocal, , the desired answer is leads to answer
Multiplying by gives
Thus, the answer is
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