ParthKohli
  • ParthKohli
Nice problem if you see it.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TheSmartOne
  • TheSmartOne
*
ParthKohli
  • ParthKohli
\[\tan n \theta = \frac{\binom{n}{1}t - \binom{n}{3}t^3 + \binom{n}5 t^5 + \cdots }{\binom{n}0 - \binom{n}2t^2 + \binom{n}4 t^4 + \cdots }\]where \(t = \tan \theta \).
anonymous
  • anonymous
Woah...

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anonymous
  • anonymous
agree ...and wat do u haveto solve for?
ParthKohli
  • ParthKohli
believe me, it's really obvious if you see the solution.
FLVSKidd
  • FLVSKidd
n6 n7 t6 t7
ganeshie8
  • ganeshie8
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
  • ganeshie8
next, \(\tan(n\theta) = \dfrac{\sin(n\theta)}{\cos(n\theta)}\), and simplifying should do...
ParthKohli
  • ParthKohli
yeah haha
ganeshie8
  • ganeshie8
to reach your formula, i think we need to divide top and bottom by \(\cos^n\theta\)

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