Waste less time on Facebook — follow Brilliant.
×

Very interesting problem

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

Note by Kiran Patel
3 years, 9 months ago

No vote yet
3 votes

Comments

Sort by:

Top Newest

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:

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

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

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

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

\(tan \frac {\pi}{11} + 4sin \frac {3\pi}{11}\)= \(\sqrt 11\) Shourya Pandey · 3 years, 9 months ago

Log in to reply

@Shourya Pandey 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. :) Bob Krueger · 3 years, 9 months ago

Log in to reply

But in my book its tan and not sine....but yes it must be sine..... Kiran Patel · 3 years, 9 months ago

Log in to reply

@Kiran Patel maybe some mistake Shourya Pandey · 3 years, 9 months ago

Log in to reply

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? Mursalin Habib · 3 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...