Nice problem if you see it.

Question

thesmartone · Answer

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parthkohli · Answer

$$\tan n \theta = \frac{\binom{n}{1}t - \binom{n}{3}t^3 + \binom{n}5 t^5 + \cdots  }{\binom{n}0 - \binom{n}2t^2 + \binom{n}4 t^4 + \cdots  }$$where $t = \tan \theta $.

anonymous · Answer

Woah...

anonymous · Answer

agree ...and wat do u haveto solve for?

parthkohli · Answer

believe me, it's really obvious if you see the solution.

flvskidd · Answer

n6 n7 t6 t7

ganeshie8 · Answer

The formulas for $\cos(n\theta)$ and $\sin(n\theta)$ follow trivially from de moivre's theorem :
$$\cos(n\theta)+i\sin(n\theta)=(\cos\theta+i\sin\theta)^n $$

expand right hand side and compare real and imaginary parts to get

ganeshie8 · Answer

next, $\tan(n\theta) = \dfrac{\sin(n\theta)}{\cos(n\theta)}$, and simplifying should do...

parthkohli · Answer

yeah haha

ganeshie8 · Answer

to reach your formula, i think we need to divide top and bottom by $\cos^n\theta$