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> | + | <center><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)}\\ | ||
Revision as of 11:32, 15 March 2015
Problem
The sum
![\[\sum_{x=2}^{44} 2\sin{x}\sin{1}[1 + \sec (x-1) \sec (x+1)]\]](http://latex.artofproblemsolving.com/a/1/5/a15e67aff38e80cc2fc3d8342336969de1100e11.png)
can be written in the form 
, where 
are trigonometric functions and 
are degrees ![$\in [0,45]$](http://latex.artofproblemsolving.com/0/c/a/0ca2266b4c86aaf1dfb58c795f59f20e52ca9974.png)
. Find 
.
Solution
By the product-to-sum identities, we know that 
, so 
:
&= \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)$ (Error compiling LaTeX. Unknown error_msg)This sum telescopes (in other words, when we expand the sum, all of the intermediate terms will cancel) to ![\[-\frac{\sin^2(1)}{\cos(1)} -\frac{\sin^2(2)}{\cos(2)} + \frac{\sin^2(44)}{\cos(44)} + \frac{\sin^2(45)}{\cos(45)}.\]](http://latex.artofproblemsolving.com/2/0/2/2028c473d3a94ea896811e9efeefe0cc8f912db2.png)
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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
