Trinometry question(prove RHS=LHS)

[MiNDTRyX]

1. The problem statement, all variables and given/known data
tan(3pi/11) + 4(sin(2pi/11) = root(11)
(umm i dont know how to do symbols sorry)
pi=3.141..... the number
and root = under-root(to the power of 1/2)
2. Relevant equations
umm i have no clue what to put here

3. The attempt at a solution
Ya see this is the thing
I don't want you guys to solve it(for now at least till I try it)
but i have absolutely no clue on how to start for this thing can someone please point me in the right direction)
Thanks

[SammyS]

I'm not sure if the following will help or not.

\tan(3x) =\frac{\sin(6x)}{\cos(6x)+1}\quad\to\quad\tan(3\pi/11) =\frac{\sin(6\pi/11)}{\cos(6\pi/11)+1}

\sin\left(\frac{6\pi}{11}\right)=\cos\left(\frac{\ pi}{2}-\frac{6\pi}{11}\right)=\cos\left(\frac{\pi}{22}\ri ght)

Similarly, \cos\left(\frac{6\pi}{11}\right)=\sin\left(\frac{\ pi}{22}\right)


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[hunt_mat]

Your gerring a number, so my feelings tell me that you will get something like a sin^2+cos^2 of something.

[SammyS]

√(11) > 1 so it's more involved than simply sin2(a) + cos2(a)

The trigonometric expression OP is asked to prove is mentioned by Eric Weisstein in the following link (http://mathworld.wolfram.com/TrigonometryAnglesPi11.html). (Expression (13), near the bottom.)

Added in Edit: Expression (12) is even more interesting. Also see (10) & (11).

[hunt_mat]

There is a coefficient of 4 in there. I am not saying that it will be easy to get to...

[SammyS]

Going a bit further:

\tan\left(\frac{3\pi}{11}\right)
=\frac{\cos\left(\frac{\pi}{22}\right)}{1+\sin\lef t(\frac{\pi}{22}\right)}
=\frac{\sqrt{\frac{1-\cos\left(\frac{\pi}{11}\right)}{2}}}{1+\sqrt{\fra c{1+\cos\left(\frac{\pi}{11}\right)}{2}}}
=\frac{\sqrt{1-\cos\left(\frac{\pi}{11}\right)}}{\sqrt{2}+\sqrt{1 +\cos\left(\frac{\pi}{11}\right)}}

This gets the tangent portion into a form with arguments of π/11 .

The other portion is: 4\sin\left(\frac{2\pi}{11}\right)=8\sin\left(\frac {\pi}{11}\right)\cos\left(\frac{\pi}{11}\right)


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[SammyS]

I found the following on Wikipedia, immediately before the "Computing π (http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Computing_.CF.80)" section:\prod_{k=1}^{m} \tan\left(\frac{k\pi}{2m+1}\right) = \sqrt{2m+1}

Which implies that for m = 5, we have:\tan\left(\frac{\pi}{11}\right)\cdot \tan\left(\frac{2\pi}{11}\right)\cdot \tan\left(\frac{3\pi}{11}\right)\cdot \tan\left(\frac{4\pi}{11}\right)\cdot \tan\left(\frac{5\pi}{11}\right)=\sqrt{11}
So, prove the above identity, and then show that:
\tan\left(\frac{\pi}{11}\right)\cdot \tan\left(\frac{2\pi}{11}\right)\cdot \tan\left(\frac{3\pi}{11}\right)\cdot \tan\left(\frac{4\pi}{11}\right)\cdot \tan\left(\frac{5\pi}{11}\right)=\tan\left(\frac{3 \pi}{11}\right)+\,4\sin\left(\frac{2\pi}{11}\right )

Added in edit;

This is equivalent to:
\tan\left(\frac{\pi}{11}\right)\cdot \tan\left(\frac{2\pi}{11}\right)\cdot \tan\left(\frac{4\pi}{11}\right)\cdot \tan\left(\frac{5\pi}{11}\right)=1+\,\frac{4\sin\l eft(\frac{2\pi}{11}\right)}{\tan\left(\frac{3\pi}{ 11}\right)}

It seems promising to write each tan function on the left as sin/cos, then use product to sum identities & make use of symmetry.

[coolul007]

An approach might be that:

\frac{3\pi}{11}= \frac{2\pi}{11} + \frac{\pi}{11}

and apply trig identities from there.