Draft:CaptainBlacks PoW #2 for MathHelpBoards

Posted on MathHelpBoards

This problem I think is a bit tedious and can be hard work. It comes from the Purdue Maths Dept PoW, only slightly modified.

Given that \(\cos(36^\circ)=\frac{1}{4}(1+\sqrt{5}) \) find \( \tan^2(18^\circ)\, \tan^2(54^\circ) \)

=============================================================

Solution 1 (this is mine and I must admit I used Maxima to handle the algebra):

First observe that from the quadrant corresponding to te angles here:

\[\tan(A/2)=\sqrt{\frac{1-\cos(A)}{1+\cos(A)}}\]

and:

\[ \tan(3A/2)=\frac{3\tan(A/2)-\tan^3(A/2)}{1-3\tan^2(A/2)} \]

So if we put \( A=36^\circ \) and allowing Maxima to do the algebra we get:

\[  \tan^2(18^\circ)\, \tan^2(54^\circ)=\frac{1}{5} \]

--------------------------------------------------------------------

Solution 2 (this is the solution give on the originating site):

\[ \cos(72^\circ)=2\cos^2(36^\circ)-1=\frac{1}{4}(\sqrt{5}-1) \]

and:

\(\displaystyle \phantom{xxxx} \tan^2(18^\circ)\, \tan^2(54^\circ) =\frac{\sin^2(54^\circ)\sin^2(18^\circ)}{\cos^2(54^\circ)\cos^2(18^\circ)} =\left[ \frac{(1/2)(\cos(36^\circ)-\cos(72^\circ))}{(1/2)(\cos(36^\circ)+\cos(72^\circ))} \right]^2\)

\(\displaystyle \phantom{xxxx} \phantom{\tan^2(18^\circ)\, \tan^2(54^\circ)}=\left[ \frac{\frac{\sqrt{5}+1}{4} -\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{5}+1}{4}+\frac{\sqrt{5}-1}{4}} \right] =\frac{1}{5}\)