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Precalculus
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verify the identity. show all work. sin^4theta-cos^4theta=sin^2theta-cos^2theta
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have u tried this ?
no idk how to start
write sin^4 theta = (sin^2 theta )^2 same for cos^4 theta what u get ?
sin ^2 theta^2 -cos^theta^2
yesh, now use \(a^2-b^2=(a+b)(a-b)\) with a=sin^2 theta, b= cos^2 theta
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so sin^theta + cos^2theta(sin^2theta-cos^2theta)
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) = ?
theta right ? or is it 1
\(\huge \color{red}{sin^2x+cos^2x = 1}\)
so its 1, so what remains ?
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sin ^2 theta -cos^2theta which is equal to the rhs (:
yes, thats it :) clear with all steps ?
yea thank you so much ur a great help
welcome ^_^
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