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How can I simplify this as much as possible?

Hi, I came across this question in an admissions test quite some time ago and didn't get it. Can someone please explain how you could simplify this expression? Thanks!
sin2(x)1+sin2(x)+sin4(x)+sin6(x)...\frac{sin^2(x)}{1 + sin^2(x)+sin^4(x)+sin^6(x)...}
Reply 1
Avatar for Zacken
Zacken
22
Original post by Glavien
Hi, I came across this question in an admissions test quite some time ago and didn't get it. Can someone please explain how you could simplify this expression? Thanks!
sin2(x)1+sin2(x)+sin4(x)+sin6(x)...\frac{sin^2(x)}{1 + sin^2(x)+sin^4(x)+sin^6(x)...}


The denominator looks like a geometric series with common ratio sin^2 x, so...?
Original post by Glavien
Hi, I came across this question in an admissions test quite some time ago and didn't get it. Can someone please explain how you could simplify this expression? Thanks!
sin2(x)1+sin2(x)+sin4(x)+sin6(x)...\frac{sin^2(x)}{1 + sin^2(x)+sin^4(x)+sin^6(x)...}


Does the bottom remind you of anything?
Reply 3
Avatar for Glavien
Glavien
OP
11
Original post by Zacken
The denominator looks like a geometric series with common ratio sin^2 x, so...?


Umm, I can't work out the sum of the series as the number of terms is not given. So, could you just work out the sum to infinity?? I am not sure why you would want to do that though.
Reply 4
Avatar for Zacken
Zacken
22
Original post by Glavien
Umm, I can't work out the sum of the series as the number of terms is not given. So, could you just work out the sum to infinity?? I am not sure why you would want to do that though.


Because there are an infinite number of terms, hence the +⋯+\cdots. Had it been +⋯sin⁡nx+\cdots \sin^n x, then that is n terms. But because it is just 3 dots, it means there are an infinite number of them.
Reply 5
Avatar for Glavien
Glavien
OP
11
Original post by Zacken
Because there are an infinite number of terms, hence the +⋯+\cdots. Had it been +⋯sin⁡nx+\cdots \sin^n x, then that is n terms. But because it is just 3 dots, it means there are an infinite number of them.


So, you work out the sum to infinity them simplify to get (sin2x)(cos2x)(sin^2x)(cos^2x)?
Reply 6
Avatar for Zacken
Zacken
22
Original post by Glavien
So, you work out the sum to infinity them simplify to get (sin2x)(cos2x)(sin^2x)(cos^2x)?


Yeah, looks right.
Reply 7
Avatar for Glavien
Glavien
OP
11
Original post by Zacken
Yeah, looks right.


Thanks, your help is really appreciated!
Reply 8
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Zacken
22
Original post by Glavien
Thanks, your help is really appreciated!


No worries. :-)

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