Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 15"
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== Solution == | == Solution == | ||
By the [[trigonometric identity|product-to-sum identities]], we know that <math>2\sin a \sin b = \cos(a-b) - \cos(a+b)</math>, so <math>2\sin{x}\sin{1} = \cos(x-1)-\cos(x+1)</math>: | By the [[trigonometric identity|product-to-sum identities]], we know that <math>2\sin a \sin b = \cos(a-b) - \cos(a+b)</math>, so <math>2\sin{x}\sin{1} = \cos(x-1)-\cos(x+1)</math>: | ||
− | + | <math> \sum_{x=2}^{44} [\cos(x-1) - \cos(x+1)][1 + \sec (x-1) \sec (x+1)]\\ | |
− | + | =\sum_{x=2}^{44} \cos(x-1) - \cos(x+1) + \frac{1}{\cos(x+1)} - \frac{1}{\cos(x-1)}\\ | |
− | + | =\sum_{x=2}^{44} \frac{\cos^2(x-1)-1}{\cos(x-1)} - \frac{\cos^2(x+1)-1}{\cos(x+1)}\\ | |
− | + | =\sum_{x=2}^{44} \left(\frac{\sin^2(x+1)}{\cos(x+1)}\right) - \left(\frac{\sin^2(x-1)}{\cos(x-1)}\right)</math> | |
− | |||
− | We now have the desired four terms. There are a couple of ways to express <math>\Phi,\,\Psi</math> as primitive trigonometric functions; for example, if we move a <math>\sin</math> to the denominator, we could express it as <math>\Phi(x) = \sin(x),\, \Psi(x) = \cot(x)</math>. Either way, we have <math>\{\theta_1,\theta_2,\theta_3,\theta_4\} = \{1^{\circ},2^{\circ},44^{\circ},45^{\circ}\}</math>, and the answer is <math>1+2+44+45 = \boxed{092}</math>. | + | This sum [[telescope]]s (in other words, when we expand the sum, all of the intermediate terms will cancel) to <math>-\frac{\sin^2(1)}{\cos(1)} -\frac{\sin^2(2)}{\cos(2)} + \frac{\sin^2(44)}{\cos(44)} + \frac{\sin^2(45)}{\cos(45)}</math>. We now have the desired four terms. There are a couple of ways to express <math>\Phi,\,\Psi</math> as primitive trigonometric functions; for example, if we move a <math>\sin</math> to the denominator, we could express it as <math>\Phi(x) = \sin(x),\, \Psi(x) = \cot(x)</math>. Either way, we have <math>\{\theta_1,\theta_2,\theta_3,\theta_4\} = \{1^{\circ},2^{\circ},44^{\circ},45^{\circ}\}</math>, and the answer is <math>1+2+44+45 = \boxed{092}</math>. |
== See also == | == See also == |
Latest revision as of 15:27, 26 December 2015
Problem
The sum can be written in the form , where are trigonometric functions and are degrees . Find .
Solution
By the product-to-sum identities, we know that , so :
This sum telescopes (in other words, when we expand the sum, all of the intermediate terms will cancel) to . We now have the desired four terms. There are a couple of ways to express as primitive trigonometric functions; for example, if we move a to the denominator, we could express it as . Either way, we have , and the answer is .
See also
Mock AIME 1 2007-2008 (Problems, Source) | ||
Preceded by Problem 14 |
Followed by Last problem | |
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