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. :)

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?

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## Comments

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TopNewestThe 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:

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

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

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

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

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

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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. :)

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

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maybe some mistake

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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

notequal to the square root of \(11\). Am I somehow misunderstanding your problem?Log in to reply