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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>:
−
<center><math> \sum_{x=2}^{44} [\cos(x-1) - \cos(x+1)][1 + \sec (x-1) \sec (x+1)]
+
<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} \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} \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></center>
+
=\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>
−
This sum [[telescope]]s (in other words, when we expand the sum, all of the intermediate terms will cancel) to <cmath>-\frac{\sin^2(1)}{\cos(1)} -\frac{\sin^2(2)}{\cos(2)} + \frac{\sin^2(44)}{\cos(44)} + \frac{\sin^2(45)}{\cos(45)}.</cmath>
 
  
−
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 \[\sum_{x=2}^{44} 2\sin{x}\sin{1}[1 + \sec (x-1) \sec (x+1)]\] can be written in the form $\sum_{n=1}^{4} (-1)^n \frac{\Phi(\theta_n)}{\Psi(\theta_n)}$, where $\Phi,\, \Psi$ are trigonometric functions and $\theta_1,\, \theta_2, \, \theta_3, \, \theta_4$ are degrees $\in [0,45]$. Find $\theta_1 + \theta_2 + \theta_3 + \theta_4$.

Solution

By the product-to-sum identities, we know that $2\sin a \sin b = \cos(a-b) - \cos(a+b)$, so $2\sin{x}\sin{1} = \cos(x-1)-\cos(x+1)$: $\sum_{x=2}^{44} [\cos(x-1) - \cos(x+1)][1 + \sec (x-1) \sec (x+1)]\\ =\sum_{x=2}^{44} \cos(x-1) - \cos(x+1) + \frac{1}{\cos(x+1)} - \frac{1}{\cos(x-1)}\\ =\sum_{x=2}^{44} \frac{\cos^2(x-1)-1}{\cos(x-1)} - \frac{\cos^2(x+1)-1}{\cos(x+1)}\\ =\sum_{x=2}^{44} \left(\frac{\sin^2(x+1)}{\cos(x+1)}\right) - \left(\frac{\sin^2(x-1)}{\cos(x-1)}\right)$

This sum telescopes (in other words, when we expand the sum, all of the intermediate terms will cancel) to $-\frac{\sin^2(1)}{\cos(1)} -\frac{\sin^2(2)}{\cos(2)} + \frac{\sin^2(44)}{\cos(44)} + \frac{\sin^2(45)}{\cos(45)}$. We now have the desired four terms. There are a couple of ways to express $\Phi,\,\Psi$ as primitive trigonometric functions; for example, if we move a $\sin$ to the denominator, we could express it as $\Phi(x) = \sin(x),\, \Psi(x) = \cot(x)$. Either way, we have $\{\theta_1,\theta_2,\theta_3,\theta_4\} = \{1^{\circ},2^{\circ},44^{\circ},45^{\circ}\}$, and the answer is $1+2+44+45 = \boxed{092}$.

See also

Mock AIME 1 2007-2008 (Problems, Source)
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