Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 6"

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== Problem ==
 
== Problem ==
If <math>\displaystyle \tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>\displaystyle 0^\circ \le \theta \le 180^\circ, </math> find <math>\displaystyle \theta.</math>
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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==
 
==Solution==
{{solution}}
 
  
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We know from product to sum formulas we have:
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<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>
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Multiply by <math>\frac{2}{2}</math>:
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<cmath>\frac{2\sin 15^\circ\cos 10^\circ-\sin 15^\circ}{\cos 15^\circ+2\cos 15^\circ\cos 10^\circ}</cmath>
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Again use product to sum:
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<cmath>\frac{\sin 5^\circ-\sin 15^\circ+\sin 25^\circ}{\cos 5^\circ+\cos 15^\circ+\cos 25^\circ}</cmath>
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Finally, use sum to product on the rightmost terms in the numerator and denominator:
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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>
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Thus, <math>\theta=\boxed{005}</math>.
 
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Revision as of 23:24, 31 May 2011

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, $\theta=\boxed{005}$.


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