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Prove that $$\tan\frac{\pi}{9}+ 4\sin\frac{\pi}{9}= \sqrt{3}$$

I think the best solution here is using right triangle . . . I have one too, but not pretty.

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Multiplying by $\cos(\pi/9)$, this is equivalent to $$\sin\frac\pi 9+2\sin\frac{2\pi}9=\sqrt3\cos\frac\pi 9.\tag1$$ But $$\sqrt3\cos\frac\pi 9-\sin\frac\pi9= 2\left(\sin\frac\pi3\cos\frac\pi 9-\cos\frac\pi3\sin\frac\pi9\right) =2\sin\left(\frac\pi3-\frac\pi9\right)=2\sin\frac{2\pi}9$$ which proves $(1)$.

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If $\cos2x\ne0,$

$$\tan2x+4\sin(30^\circ-x)$$ $$=\dfrac{\sin2x+4\sin(30^\circ-x)\cos(30^\circ-x)}{\cos2x}$$

$$=\dfrac{\sin2x+2\sin(60^\circ-2x)}{\cos2x}$$

$$=\dfrac{\sin2x+\sqrt3\cos2x-\sin2x}{\cos2x}=?$$

Here $x=10^\circ$

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$$(\tan x+ 4\sin x)^{2}- 3= - \frac{1}{4}(2\cos x+ 1)(2\cos 3x- 1)\csc^{2}\left ( \frac{\pi}{4}- \frac{x}{2} \right )\csc^{2}\left ( \frac{x}{2}+ \frac{\pi}{4} \right )$$ Therefore $$\tan\frac{\pi}{9}+ 4\sin\frac{\pi}{9}= \sqrt{3}$$

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  • $\begingroup$ .solve: @Eynjuhl $\endgroup$ – user707745 Sep 26 '19 at 6:03

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