Question

Nice problem if you see it.

Answer 1

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Answer 2

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

Answer 3

Woah...

Answer 4

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

Answer 5

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

Answer 6

n6 n7 t6 t7

Answer 7

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

Answer 8

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

Answer 9

yeah haha

Answer 10

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