Difference between revisions of "2012 AIME II Problems/Problem 9"
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So, <math>\sin^2 y = \frac{3}{35}</math>. | So, <math>\sin^2 y = \frac{3}{35}</math>. | ||
− | Squaring Equation 1 (leading to <math>\sin^2 x = 9\sin^2 y</math>), we can solve for <math>\sin^2 | + | Squaring Equation 1 (leading to <math>\sin^2 x = 9\sin^2 y</math>), we can solve for <math>\sin^2 x</math>: |
<math>\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}</math> | <math>\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}</math> |
Revision as of 23:41, 7 July 2012
Contents
Problem 9
Let and be real numbers such that and . The value of can be expressed in the form , where and are relatively prime positive integers. Find .
Solution
Examine the first term in the expression we want to evaluate, , separately from the second term, .
The First Term
Using the identity , we have:
The Second Term
Let the equation be equation 1, and let the equation be equation 2. Hungry for the widely-used identity , we cross multiply equation 1 by and multiply equation 2 by .
Equation 1 then becomes:
.
Equation 2 then becomes:
Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:
Applying the identity (which is similar to but a bit different), we can change into:
Rearranging, we get .
So, .
Squaring Equation 1 (leading to ), we can solve for :
Using the identity , we can solve for .
Thus, .
Now Back to the Solution!
Finally, .
So, the answer is .
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |