# Very interesting problem

Prove tan{(3pi)/11} + 4tan{(2pi)/11} =$11^{0.5}$= square root of 11...

Note by Kiran Patel
4 years, 11 months ago

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The problem actually is as follows, and I request the user to edit it. Iam not giving the solution, and that googling it will give the solution:

Prove that $$tan \frac {3\pi}{11} + 4sin \frac {2\pi}{11}$$= $$\sqrt 11$$

Infact, even the following relations hold:

$$tan \frac {3\pi}{11} + 4sin \frac {2\pi}{11}$$= $$\sqrt 11$$

$$tan \frac {4\pi}{11} + 4sin \frac {\pi}{11}$$= $$\sqrt 11$$

$$tan \frac {5\pi}{11} - 4sin \frac {4\pi}{11}$$= $$\sqrt 11$$

$$tan \frac {2\pi}{11} - 4sin \frac {5\pi}{11}$$= $$-\sqrt 11$$

$$tan \frac {\pi}{11} + 4sin \frac {3\pi}{11}$$= $$\sqrt 11$$

- 4 years, 11 months ago

By the way, you should put braces around your eleven's that are under the square root, so that the entire radican is covered by the bar of the square root. It should look like \sqrt{11} rather than \sqrt11, so that $$\sqrt{11}$$, rather than $$\sqrt11$$, appears. :)

- 4 years, 11 months ago

But in my book its tan and not sine....but yes it must be sine.....

- 4 years, 11 months ago

maybe some mistake

- 4 years, 11 months ago

Are you sure this is what we have to prove? Because I just used a calculator and saw that $$tan(\frac{3\pi}{11}) +4tan(\frac{2\pi}{11})$$ is not equal to the square root of $$11$$. Am I somehow misunderstanding your problem?

- 4 years, 11 months ago