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

Question

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.

Answer 1

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 $(1)$.

Answer 2

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 $x=10^\circ$

Answer 3

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

Answer 4

Here is an easy approach via complex numbers.

Recall that the roots to $x^n-1$ are $1,u,u^2,\dots,u^{n-1}$ where $u=e^{\frac{2i\pi}{n}}$. Also recall that via Euler's identity $\sin \theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$ and $\tan \theta = \frac{e^{i\theta}-e^{-i\theta}}{i(e^{i\theta}+e^{-i\theta})}$.

Assume $u = e^{\frac{2i\pi}{9}}$. It follows that $u^9=1$ and $u^{\frac{9}{2}} = -1$.

Also $4i\sin \frac{\pi}{9} = 2(u^{\frac{1}{2}} - u^{-\frac{1}{2}})$

But $u^{\frac{1}{2}} = -u^5$ and $u^{-\frac{1}{2}} = -u^4$.

So $4i\sin \frac{\pi}{9} = 2(u^4-u^5)$

And $i\tan \frac{\pi}{9} = \frac{u-1}{u+1} = \frac{u-u^9}{u+1} = u+u^3+u^5+u^7-u^2-u^4-u^6-u^8$.

So, $$i\tan \frac{\pi}{9} +4i\sin \frac{\pi}{9} = \underbrace{u+u^3+u^4+u^7}_a-(\underbrace{u^2+u^5+u^6+u^8}_b)$$

We know $a+b=-1$ and $ab$ can be easily calculated as $3+2(u^3+u^6) = 1$. So $$(a-b)^2 = (a+b)^2-4ab = -3 \implies a-b = \pm i\sqrt{3}$$

Since $\frac{\pi}{9}$ lies in the first quadrant, it follows that all its trigonometric functions are positive.

So $$a-b=i\tan \frac{\pi}{9}+4i\sin \frac{\pi}{9} = i\sqrt{3}$$

Quadratic Gauss sums may also be of interest as a generalization to find $a-b$.