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Calculate $\tan9^{\circ}-\tan27^{\circ}-\tan63^{\circ}+\tan81^{\circ}$?

The correct answer should be 4.

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You'd better to know $$\sin 2\theta = 2\sin \theta \cos \theta$$ $$\sin \alpha-\sin \beta= 2\cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$$

Then we have \begin{align*} \tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ} &= \frac{\sin 9^{\circ}}{\cos 9^{\circ}}- \frac{\sin 27^{\circ}}{\cos 27^{\circ}}- \frac{\cos 27^{\circ}}{\sin 27^{\circ}}+ \frac{\cos 9^{\circ}}{\sin 9^{\circ}} \\ &= \frac{1}{\sin 9^{\circ} \cos 9^{\circ}}- \frac{1}{\sin 27^{\circ} \cos 27^{\circ}} \\ &= \frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}} \\ &= \frac{2(\sin 54^{\circ}-\sin 18^{\circ})}{\sin 18^{\circ} \sin 54^{\circ}} \\ &= 2\frac{2\cos 36^{\circ} \sin 18^{\circ}}{\sin 18^{\circ} \cos 36^{\circ}} \\ &= 4 \end{align*}

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  • $\begingroup$ Okay, I have found it. Cheers! $\endgroup$ – 52145208 Jun 7 '14 at 7:03
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$\tan9^{\circ} + \tan81^{\circ} = \dfrac{\sin9^{\circ}}{\cos9^{\circ}} + \dfrac{\sin81^{\circ}}{\cos81^{\circ}} = \dfrac{\sin(9^{\circ} + 81^{\circ})}{\cos9^{\circ}\cos81^{\circ}} = \dfrac{1}{\cos9^{\circ}\cos81^{\circ}} = \dfrac{1}{\cos9^{\circ}\sin9^{\circ}} = \dfrac{2}{\sin18^{\circ}}$. Similarly:

$\tan27^{\circ} + \tan63^{\circ} = \dfrac{1}{\cos27^{\circ}\cos63^{\circ}} = \dfrac{1}{\cos27^{\circ}\sin27^{\circ}} = \dfrac{2}{\sin54^{\circ}}$.

Now let $x = \sin18^{\circ}$, then from $\cos36^{\circ} = \sin54^{\circ}$ we have:

$1 - 2x^2 = 3x - 4x^3 \to 4x^3 - 2x^2 - 3x + 1 = 0 \to (x-1)(4x^2 + 2x - 1) = 0$, since $x \neq 1$, $x = \dfrac{\sqrt{5} - 1}{4} = \sin18^{\circ}$, and $\sin54^{\circ} = 1 - 2x^2 = 1 - 2\left(\dfrac{\sqrt{5} - 1}{4}\right)^2 = \dfrac{\sqrt{5} + 1}{4}$. Thus:

$S = \dfrac{2}{\dfrac{\sqrt{5} - 1}{4}} -\dfrac{2}{\dfrac{\sqrt{5} + 1}{4}} = 4$

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  • $\begingroup$ This is very helpful! Cheers! $\endgroup$ – 52145208 Jun 7 '14 at 7:09
  • $\begingroup$ @8pir, How have you discarded the other root of the Quadratic Equation $\endgroup$ – lab bhattacharjee Jun 7 '14 at 7:41
  • $\begingroup$ @labbhattacharjee: the other root is: $\dfrac{-1 - \sqrt{5}}{4} < 0$ $\endgroup$ – DeepSea Jun 7 '14 at 7:54
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    $\begingroup$ @8pir, That should be part of the answer. You have basically derived the value of $\sin18^\circ$. Would you mind looking into my answer? $\endgroup$ – lab bhattacharjee Jun 7 '14 at 7:59
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Observe that $\displaystyle\tan(5\cdot27^\circ)=\tan(180^\circ-45^\circ)=-\tan45^\circ=-1$

$\displaystyle\implies\tan(-27^\circ\cdot5)=-\tan(5\cdot27^\circ)=1$

Similarly we can show that $\displaystyle\tan5x=1$ for $\displaystyle x=-27^\circ,-63^\circ,9^\circ,81^\circ,45^\circ$

From Sum of tangent functions where arguments are in specific arithmetic series, $$\tan5x=\frac{\binom51t-\binom53t^3+t^5}{1-\binom52t^2+\binom54t^4}\text{ where } t=\tan x$$

$$5x=45^\circ\implies t^5-5t^4\tan45^\circ+\cdots=0$$ has the roots $\displaystyle \tan(-27^\circ),\tan(-63^\circ),\tan9^\circ,\tan81^\circ,\tan45^\circ$

Using Vieta's formula, $\displaystyle\tan(-27^\circ)+\tan(-63^\circ)+\tan9^\circ+\tan81^\circ+\tan45^\circ=\frac51$

See also : Prove the trigonometric identity $(35)$

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  • $\begingroup$ To be honest, I cannot quite understand. Anyway, thanks a lot! I have to read it deeply. $\endgroup$ – 52145208 Jun 7 '14 at 8:40
  • $\begingroup$ @52145208, Really sorry to learn that.Have you heard Vieta's formulae? Where you are actually stuck? $\endgroup$ – lab bhattacharjee Jun 7 '14 at 8:48
  • $\begingroup$ Okay, I think I understand! This way is so cool! Thank you! $\endgroup$ – 52145208 Jun 7 '14 at 9:19

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