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Precalculus 18 Online
OpenStudy (anonymous):

verify the identity. show all work. sin^4theta-cos^4theta=sin^2theta-cos^2theta

hartnn (hartnn):

have u tried this ?

OpenStudy (anonymous):

no idk how to start

hartnn (hartnn):

write sin^4 theta = (sin^2 theta )^2 same for cos^4 theta what u get ?

OpenStudy (anonymous):

sin ^2 theta^2 -cos^theta^2

hartnn (hartnn):

yesh, now use \(a^2-b^2=(a+b)(a-b)\) with a=sin^2 theta, b= cos^2 theta

OpenStudy (anonymous):

so sin^theta + cos^2theta(sin^2theta-cos^2theta)

hartnn (hartnn):

yeah, don't miss out brackets , (sin^2 theta + cos^2 theta)(sin^2theta-cos^2theta) and i think you know what is (sin^2 theta + cos^2 theta) = ?

OpenStudy (anonymous):

theta right ? or is it 1

hartnn (hartnn):

\(\huge \color{red}{sin^2x+cos^2x = 1}\)

hartnn (hartnn):

so its 1, so what remains ?

OpenStudy (anonymous):

sin ^2 theta -cos^2theta which is equal to the rhs (:

hartnn (hartnn):

yes, thats it :) clear with all steps ?

OpenStudy (anonymous):

yea thank you so much ur a great help

hartnn (hartnn):

welcome ^_^

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