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Trigonometry 16 Online
OpenStudy (anonymous):

Find the identical expression for 1-sin^4theta/1+sin^2theta

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 ?

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

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 .

OpenStudy (anonymous):

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