MHB CaptainBlacks Problem of the Week #2

[CaptainBlack]

I think this is a bit tedious and can be hard labour, but:

Given that \(\cos(36^\circ )=\frac{1}{4}(1+\sqrt{5})\) find the exact value of \(\tan^2(18^\circ) \tan^2(54^\circ)\).

CB

 

[Sudharaka]

I think this is a bit tedious and can be hard labour, but:

Given that \(\cos(36^\circ )=\frac{1}{4}(1+\sqrt{5})\) find the exact value of \(\tan^2(18^\circ) \tan^2(54^\circ)\).

CB

\[\cos 2\theta=2\cos^{2}\theta-1\Rightarrow\cos^{2}\theta=\frac{1+\cos 2\theta}{2}~~~~~(1)\]

\[\cos 2\theta=2\cos^{2}\theta-1\Rightarrow\cos^{2}\theta=\frac{1+\cos 2\theta}{2}~~~~~(2)\]

By (1) and (2),

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

Also by the Triple angle formula for tangents,

\[\tan 3\theta = \frac{3 \tan\theta - \tan^3\theta}{1 - 3 \tan^2\theta}\]

Substituting \(\tan\theta=\sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}}\) and simplifying yields,

\[\tan 3\theta=\pm\frac{\sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta}}\left(\frac{2\cos 2\theta+1}{2\cos 2\theta-1}\right)~~~~~(B)\]

By (A) and (B),

\[\tan^{2}\theta\tan^{2}3\theta=\left(\frac{1-\cos 2\theta}{1+\cos 2\theta}\right)^2\left(\frac{2\cos 2\theta+1}{2\cos 2\theta-1}\right)^2\]

Let \(\theta=18^{0}\),

\[\tan^{2}(18^{0})\tan^{2}(54^{0})=\left(\frac{1-\cos (36^{0})}{1+\cos (36^{0})}\right)^2\left(\frac{2\cos (36^{0})+1}{2\cos (36^{0})-1}\right)^2\]

Since \(\cos (36^{0})=\frac{1}{4}(1+\sqrt{5})\),

\[\tan^{2}(18^{0})\tan^{2}(54^{0})=\left(\frac{3-\sqrt{5}}{5+\sqrt{5}}\right)^{2}\left(\frac{3+ \sqrt{5}}{\sqrt{5}-1}\right)^{2}\]

\[\Rightarrow \tan^{2}(18^{0})\tan^{2}(54^{0})=\frac{1}{5}\left(\frac{9-5}{5-1}\right)^2\]

\[\therefore \tan^{2}(18^{0})\tan^{2}(54^{0})=0.2\]

 

[CaptainBlack]

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:\(\displaystyle \phantom{xxxx} \tan^2(18^\circ)\, \tan^2(54^\circ) =\frac{\sin^2(54^\circ)\sin^2(18^\circ)}{\cos^2(54^\circ)\cos^2(18^\circ)} =\left[ \frac{(1/2)(\cos(36^\circ)-\cos(72^\circ))}{(1/2)(\cos(36^\circ)+\cos(72^\circ))} \right]^2\)\(\displaystyle \phantom{xxxx} \phantom{\tan^2(18^\circ)\, \tan^2(54^\circ)}=\left[ \frac{\frac{\sqrt{5}+1}{4} -\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{5}+1}{4}+\frac{\sqrt{5}-1}{4}} \right] =\frac{1}{5}\)