$\phi$: golden ratio, $\Phi={1\over \phi}$

I want to show that,

$${\Phi\tan{9^\circ}-\phi\tan{27^\circ}\over \sin^2{9^\circ}-\sin^2{27^\circ}}=4$$

Using $\sin^2{x}={1\over 2}(1-\cos{2x})$

$${\Phi\tan{9^o}-\phi\tan{27^o}\over \cos{54^\circ}-\cos{18^\circ}}=2$$

As for numerator

$${\sqrt5(\tan{27^\circ}-\tan{9^\circ})-\tan{27^\circ}-\tan{9^\circ}\over \cos{54^\circ}-\cos{18^\circ}}=4$$

exact trig values

if I was to plug in the corresponding values it would be a lot of work to simply and might not even get to the result.

Any idea would help me to simplify the LHS?

  • $\begingroup$ I think you simplified the numerator incorrectly, even after your edit. It should be: $$\Phi \tan(9^\circ)-\phi \tan(27^\circ)=\frac{\sqrt{5}-1}{2}\tan(9^\circ)-\frac{1+\sqrt{5}}{2}\tan(27^\circ)=\frac{\sqrt{5}(\tan(9^\circ)-\tan(27^\circ))-\tan(9^\circ)-\tan(27^\circ)}{2}$$ $\endgroup$ – Noble Mushtak Jun 22 '16 at 13:32
  • $\begingroup$ In the intermediate step, you have a factor of $\frac{1}{2}$ in the denominator, so when you multiply that factor out, you should get $4\frac 1 2=2$ on the right side. You still get $4$ in the end, but I'm going off the intermediate step in my answer, so putting the correct $=2$ instead of the incorrect $=8$ on that step is important. $\endgroup$ – Noble Mushtak Jun 22 '16 at 13:48

Noting that $$ \phi=2\cos(\frac{\pi}{5})=2\sin(\frac{3\pi}{10}), \Phi=2\sin(\frac{\pi}{10})$$ from https://en.wikipedia.org/wiki/Golden_ratio, so one has \begin{eqnarray} {\Phi\tan{9^\circ}-\phi\tan{27^\circ}\over \sin^2{9^\circ}-\sin^2{27^\circ}}&=&{2\sin(\frac{\pi}{10})\tan(\frac{\pi}{20})-2\sin(\frac{3\pi}{10})\tan(\frac{3\pi}{20})\over \sin^2(\frac{\pi}{20})-\sin^2(\frac{3\pi}{20})}\\ &=&2{2\sin(\frac{\pi}{20})\cos(\frac{\pi}{20})\tan(\frac{\pi}{20})-2\sin(\frac{3\pi}{20})\cos(\frac{3\pi}{20})\tan(\frac{3\pi}{20})\over \sin^2(\frac{\pi}{20})-\sin^2(\frac{3\pi}{20})}\\ &=&4{\sin^2(\frac{\pi}{20})-\sin^2(\frac{3\pi}{20})\over \sin^2(\frac{\pi}{20})-\sin^2(\frac{3\pi}{20})}\\ &=&4. \end{eqnarray}


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