Trinometry question(prove RHS=LHS)

[MiNDTRyX]

Homework Statement

tan(3pi/11) + 4(sin(2pi/11) = root(11)
(umm i don't know how to do symbols sorry)
pi=3.141... the number
and root = under-root(to the power of 1/2)

Homework Equations

umm i have no clue what to put here

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}\right)

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


.

 

[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 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)<br /> =\frac{\cos\left(\frac{\pi}{22}\right)}{1+\sin\left(\frac{\pi}{22}\right)}<br /> =\frac{\sqrt{\frac{1-\cos\left(\frac{\pi}{11}\right)}{2}}}{1+\sqrt{\frac{1+\cos\left(\frac{\pi}{11}\right)}{2}}}<br /> =\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).

 

[SammyS]

I found the following on Wikipedia, immediately before the "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}<br />​

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:

<br /> \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:

<br /> \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\left(\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.