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This question was asked as an equality on MSE and I am quite surprised to find that its strictly false

However I would like to see why is their difference of the order $10^{-15}$?

$$\tan 54^\circ\approx \dfrac{\sin24^\circ}{1-\sqrt{3}\sin24^\circ}$$

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    $\begingroup$ If you ask Wolfram Alpha for an explicit form for $\tan\left(\frac{3\pi}{10}\right)$ it gives $\sqrt{1+\frac{2}{\sqrt{5}}}$. If you ask it what $\frac{\sin(\frac{2\pi}{15})}{1-\sqrt{3}\sin(\frac{2\pi}{15})}$ is, it will give $\sqrt{\frac15(5+2\sqrt{5})}$ as an exact form. It just can't figure out they're equal through all the conversions, I suspect. $\endgroup$ – Steven Stadnicki Aug 7 '14 at 15:52
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    $\begingroup$ Mathematica uses (normally) floating point accuracy in the calculations, which is roughly $10^{-15}$. One can increase the accuracy by using: ' SetAccuracy[expression,n] ' where $n$ is the number of digits needed. Applying this to the equation above and taking $n=30$ gives $0$ (to an accuracy of $10^{-30}$). $\endgroup$ – Winther Aug 7 '14 at 15:56
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    $\begingroup$ I have posted a correct complete proof to the original question. $\endgroup$ – Rene Schipperus Aug 7 '14 at 18:29
  • $\begingroup$ possible duplicate of Prove $\tan 54^\circ=\frac{\sin24^\circ}{1-\sqrt{3}\sin24^\circ}$ $\endgroup$ – Thomas Produit Jan 29 '15 at 11:06
  • $\begingroup$ @Thomas The OP links to your alleged duplicate in their post - they are asking about the accuracy of the approximation, and why it is what it is. "If they are equal, then why is Wolfram-Alpha saying otherwise?" $\endgroup$ – user1729 Jan 29 '15 at 11:43
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$\tan 54^\circ\color{red}{=} \frac{\sin24^\circ}{1-\sqrt{3}\sin24^\circ}$ is true as darth geek proved analytically. The difference of $10^{-15}$ is a numerical error.

EDIT: One can prove this directly by using the exact values of sine, cosine and tangent.

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  • $\begingroup$ If you mean that it is true then prove it and not just say so! Edit: That is much better, +1! $\endgroup$ – Winther Aug 7 '14 at 15:46
  • $\begingroup$ Why doesn't Wolfram does say so?? $\endgroup$ – Tom Lynd Aug 7 '14 at 15:46
  • $\begingroup$ @metacompactness I ashall accept your answer if you prove the identity, for bonus :) $\endgroup$ – Tom Lynd Aug 7 '14 at 15:52
  • $\begingroup$ @metacompatness Can you complete it... $\endgroup$ – Tom Lynd Aug 7 '14 at 16:05
  • $\begingroup$ great lets see the proof. $\endgroup$ – Rene Schipperus Aug 7 '14 at 19:45

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