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 10:51, 4 April 2012
Problem
If and find
Solution
We know from product to sum formulas we have: Multiply by : Again use product to sum: Finally, use sum to product on the rightmost terms in the numerator and denominator: 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 |