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>.
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==See Also==
 
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{{Mock AIME box|year=2006-2007|n=2|num-b=5|num-a=7}}
*[[Mock AIME 2 2006-2007 Problems/Problem 5 | Previous Problem]]
 
 
 
*[[Mock AIME 2 2006-2007 Problems/Problem 7 | Next Problem]]
 
 
 
*[[Mock AIME 2 2006-2007]]
 

Latest revision as of 10:51, 4 April 2012

Problem

If $\tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta$ and $0^\circ \le \theta \le 180^\circ,$ find $\theta.$

Solution

We know from product to sum formulas we have: \[\frac{\sin 15^\circ\sin 25^\circ\sin 35^\circ}{\cos 15^\circ\cos 25^\circ\cos 35^\circ}=\frac{\sin 15^{\circ}(\cos 10^\circ-\cos 60^\circ)}{\cos 15^\circ(\cos 10^\circ+\cos 60^\circ)}\] Multiply by $\frac{2}{2}$: \[\frac{2\sin 15^\circ\cos 10^\circ-\sin 15^\circ}{\cos 15^\circ+2\cos 15^\circ\cos 10^\circ}\] Again use product to sum: \[\frac{\sin 5^\circ-\sin 15^\circ+\sin 25^\circ}{\cos 5^\circ+\cos 15^\circ+\cos 25^\circ}\] Finally, use sum to product on the rightmost terms in the numerator and denominator: \[\frac{\sin 5^{\circ}+2\sin 5^\circ\cos 20^\circ}{\cos 5^\circ+2\cos 5^\circ\cos 20^\circ}=\frac{\sin 5^\circ(1+2\cos 20^\circ)}{\cos 5^\circ(1+2\cos 20^\circ)}=\tan 5^\circ\] Thus, $\theta=\boxed{005}$.

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
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