Question

Simplify (sin^2(x))/(1+(sin^2(x))+(sin^4(x))+(sin^6(x))+(sin^8(x))+...)

Answer 1

\[ \Large \sum_{i=1}^n\left(\sin^2(x)\right)^{n-1}\]
Seems to be geometric:
\[\Large s_n=\frac{1-q^n}{1-q}=\frac{1-\sin^{2n}(x)}{1-\sin^2(x)}=\frac{1-\sin^{2n}(x)}{\cos^2(x)} \]

Answer 2

for the denominator of your given term.

Answer 3

The rest I would guess is a matter of art and interpretation, for large numbers of n, your sin^(2n)(x) cycles between 0 and 1, so it is possible for that term to converge, if that's the case your answer would be:

\[\Large \sin^2(x)\cos^2(x) \]

Answer 4

So written out it might be more informative:
\[\Large \sum_{n=1}^\infty\left(\sin^2(x)\right)^{n-1}=\frac{1}{\cos^2(x)}=\sec^2(x) \]

And

\[\Large \frac{\sin^2(x)}{\sum_{n=1}^\infty\left(\sin^2(x)\right)^{n-1}}=\frac{\sin^2(x)}{\sec^2(x)}=\sin^2(x)\cos^2(x) \]

Answer 5

Thanks! To add on,
Since \[2\sin(x)\cos(x)=\sin(2x)\]
Then \[\sin^2(x)\cos^2(x)=(\sin(x)\cos(x))^2=(1/2\sin(2x))^2=1/4\sin^2(2x)\]

Answer 6

Yes, very true! Makes it even more elegant. (-:

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