Simplify: (sin Θ − cos Θ)2 − (sin Θ + cos Θ)2 a. −4sin(Θ)cos(Θ) b. 2 c. sin^2 Θ d. cos^2 Θ

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Simplify: (sin Θ − cos Θ)2 − (sin Θ + cos Θ)2 a. −4sin(Θ)cos(Θ) b. 2 c. sin^2 Θ d. cos^2 Θ

Mathematics
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\[(\sin(\theta)-\cos(\theta))^2-(\sin(\theta)+\cos(\theta))^2\] Apply the identity \[a^2-b^2=(a+b)(a-b)\] \[\implies ((\sin(\theta)-\cos(\theta))+(\sin(\theta)+\cos(\theta)))\times((\sin(\theta)-\cos(\theta))-(\sin(\theta)+\cos(\theta)))\]
Here, we've taken \[a=\sin(\theta)-\cos(\theta)\] and \[b=\sin(\theta)+\cos(\theta)\]
Wouldnt the answer be 0 tho? since it's cancelled out?

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Nope, open the parenthesis and find out for yourself
i still dont understand
hint: you can use these identities: \[\Large \begin{gathered} {\left( {\sin \theta - \cos \theta } \right)^2} = {\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} - 2\sin \theta \cos \theta \hfill \\ \hfill \\ {\left( {\sin \theta + \cos \theta } \right)^2} = {\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} + 2\sin \theta \cos \theta \hfill \\ \end{gathered} \]
Hm, it seems much more simpler to understand if you apply those identities above, do you understand them @xoamb
\[(a+b)^2=a^2+b^2+2ab\] \[(a-b)^2=a^2+b^2-2ab\]
so it would be d?
Sorry, we just don't give answers, and it seems like all you're doing is just guessing, why not at least show us some work so we can understand where exactly you are having problem understanding the concept

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