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What is the simplification of $\sin^2 x/(1+ \sin^2 x +\sin^4 x +\sin^6 x + \cdots)$?

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You can enclose LaTeX in $-signs. –  t.b. Feb 9 '11 at 15:09

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up vote 2 down vote accepted

Assuming that $x \notin \frac{\pi}{2} + \pi \mathbb{Z}$ you can write $q = \sin^{2}{x}$ with $|q| < 1$ and use the geometric series $\sum_{n=0}^{\infty} q^{n} = \frac{1}{1-q}$.

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So the answer would be sin^2 (x) . ( 1-sin^2 (x) ) ?? –  user6834 Feb 9 '11 at 15:13
    
@Mr D: Yes, under the assumption about $x$ that I made. Note that this can be further simplified by using $\sin^2 x + \cos^2 x = 1$ and the formula $\sin{2x} = 2\sin{x}\cos{x}$ –  t.b. Feb 9 '11 at 15:15
    
Thank you very much –  user6834 Feb 9 '11 at 15:30
    
@Mr D: Note that you still have to consider the two cases $\sin{x} = \pm 1$ –  t.b. Feb 9 '11 at 15:32

What does $1 + \sin^2 x + \sin^4 x + \sin^6 x + ....$ simplify to?

Or better, what does $1 + x^2 + x^4 + x^6 + ....$ simplify to?

Or better, what does $1 + x + x^2 + x^3 + ....$ simplify to?

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