# CaptainBlacks Problem of the Week #2

#### CaptainBlack

##### Well-known member
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

Last edited:
• Sudharaka

#### Sudharaka

##### Well-known member
MHB Math Helper
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 $$\cos (36^{0})=\frac{1}{4}(1+\sqrt{5})$$,

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

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

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

• sbhatnagar

#### CaptainBlack

##### Well-known member
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 $$\tan^2(18^\circ)\, \tan^2(54^\circ)$$

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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:

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

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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:

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

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

• sbhatnagar and Sudharaka