Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 6"
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<cmath>\frac{\sin 5^{\circ}+2\sin 5^\circ\cos 20^\circ}{\cos 5^\circ+2\cos 5^\circ\cos 20^\circ}=\frac{\sin 5^\circ(1+2\cos 20^\circ)}{\cos 5^\circ(1+2\cos 20^\circ)}=\tan 5^\circ</cmath> | <cmath>\frac{\sin 5^{\circ}+2\sin 5^\circ\cos 20^\circ}{\cos 5^\circ+2\cos 5^\circ\cos 20^\circ}=\frac{\sin 5^\circ(1+2\cos 20^\circ)}{\cos 5^\circ(1+2\cos 20^\circ)}=\tan 5^\circ</cmath> | ||
Thus, <math>\theta=\boxed{005}</math>. | Thus, <math>\theta=\boxed{005}</math>. | ||
| − | + | ==See Also== | |
| − | + | {{Mock AIME box|year=2006-2007|n=2|num-b=5|num-a=7}} | |
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Latest revision as of 09:51, 4 April 2012
Problem
If 
and 
find 
Solution
We know from product to sum formulas we have:
![\[\frac{\sin 15^\circ\sin 25^\circ\sin 35^\circ}{\cos 15^\circ\cos 25^\circ\cos 35^\circ}=\frac{\sin 15^{\circ}(\cos 10^\circ-\cos 60^\circ)}{\cos 15^\circ(\cos 10^\circ+\cos 60^\circ)}\]](http://latex.artofproblemsolving.com/9/d/8/9d8b54c560558bb064a19b752b31e0cc1433482d.png)
Multiply by 
:
![\[\frac{2\sin 15^\circ\cos 10^\circ-\sin 15^\circ}{\cos 15^\circ+2\cos 15^\circ\cos 10^\circ}\]](http://latex.artofproblemsolving.com/0/5/a/05a9baab2f4c3765854c32b9fe6f3d95e9b3fb75.png)
Again use product to sum:
![\[\frac{\sin 5^\circ-\sin 15^\circ+\sin 25^\circ}{\cos 5^\circ+\cos 15^\circ+\cos 25^\circ}\]](http://latex.artofproblemsolving.com/a/e/9/ae9e798b4be45aab4b26ada64f678a66f7430a26.png)
Finally, use sum to product on the rightmost terms in the numerator and denominator:
![\[\frac{\sin 5^{\circ}+2\sin 5^\circ\cos 20^\circ}{\cos 5^\circ+2\cos 5^\circ\cos 20^\circ}=\frac{\sin 5^\circ(1+2\cos 20^\circ)}{\cos 5^\circ(1+2\cos 20^\circ)}=\tan 5^\circ\]](http://latex.artofproblemsolving.com/f/a/6/fa6462667ae6ae907ec0d59bea07a99ab69377a0.png)
Thus, 
.
See Also
| Mock AIME 2 2006-2007 (Problems, Source) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
