Find the identical expression for
1-sin^4theta/1+sin^2theta
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OpenStudy (anonymous):
1-sin^4theta=(1+sin^2theta)(1+sin^2theta)
so, it all simplifies to 1+sin^2theta.
OpenStudy (anonymous):
none of my answer choices are 1+sin^2theta
OpenStudy (anonymous):
plus you missed the 1+sin^2theta on the bottom
hartnn (hartnn):
he must have meant 1-sin^2theta= cos^2 theta
hartnn (hartnn):
is cos^2 theta in options ?
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OpenStudy (anonymous):
yes it is
hartnn (hartnn):
good :)
OpenStudy (anonymous):
so how exactly do I solve it then?
OpenStudy (amorfide):
1-sin^4theta
this is a difference of two squares so factorise it
(1+sin²theta)(1-sin²theta)
so now you have
(1+sin²theta)(1-sin²theta)/1+sin^2theta
this cancels out to 1-sin²theta
1-sin²theta=cos²theta
OpenStudy (amorfide):
1-sin²theta=cos²theta this comes from
sin²theta + cos²theta=1
rearrange
1-sin²theta=cos²theta
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OpenStudy (amorfide):
do you understand?
OpenStudy (anonymous):
1-sin2teta=cos2teta
OpenStudy (amorfide):
@mahmit2012 that is wrong, sin2theta has its own identity sin²theta is not sin 2 theta
OpenStudy (anonymous):
yes I think I understand, thank you!
OpenStudy (anonymous):
I wrote sin^2 !! check it out .
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