Mock AIME 1 2007-2008 Problems/Problem 15

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 $1+2+44+45 = \boxed{092}$ .

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by Problem 14	Followed by Last problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Page

Toolbox

Search

Mock AIME 1 2007-2008 Problems/Problem 15

Problem

Solution

See also