Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 6"
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== Problem == | == Problem == | ||
− | If <math> | + | If <math>\tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>0^\circ \le \theta \le 180^\circ, </math> find <math>\theta.</math> |
+ | |||
+ | ==Solution== | ||
+ | |||
+ | We know from product to sum formulas we have: | ||
+ | <cmath>\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)}</cmath> | ||
+ | Multiply by <math>\frac{2}{2}</math>: | ||
+ | <cmath>\frac{2\sin 15^\circ\cos 10^\circ-\sin 15^\circ}{\cos 15^\circ+2\cos 15^\circ\cos 10^\circ}</cmath> | ||
+ | Again use product to sum: | ||
+ | <cmath>\frac{\sin 5^\circ-\sin 15^\circ+\sin 25^\circ}{\cos 5^\circ+\cos 15^\circ+\cos 25^\circ}</cmath> | ||
+ | Finally, use sum to product on the rightmost terms in the numerator and denominator: | ||
+ | <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>. | ||
+ | ==See Also== | ||
+ | {{Mock AIME box|year=2006-2007|n=2|num-b=5|num-a=7}} |
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 |