anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[(\sin(\theta)-\cos(\theta))^2-(\sin(\theta)+\cos(\theta))^2\] Apply the identity \[a^2-b^2=(a+b)(a-b)\] \[\implies ((\sin(\theta)-\cos(\theta))+(\sin(\theta)+\cos(\theta)))\times((\sin(\theta)-\cos(\theta))-(\sin(\theta)+\cos(\theta)))\]
anonymous
  • anonymous
Here, we've taken \[a=\sin(\theta)-\cos(\theta)\] and \[b=\sin(\theta)+\cos(\theta)\]
anonymous
  • anonymous
Wouldnt the answer be 0 tho? since it's cancelled out?

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anonymous
  • anonymous
Nope, open the parenthesis and find out for yourself
anonymous
  • anonymous
i still dont understand
Michele_Laino
  • Michele_Laino
hint: you can use these identities: \[\Large \begin{gathered} {\left( {\sin \theta - \cos \theta } \right)^2} = {\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} - 2\sin \theta \cos \theta \hfill \\ \hfill \\ {\left( {\sin \theta + \cos \theta } \right)^2} = {\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} + 2\sin \theta \cos \theta \hfill \\ \end{gathered} \]
anonymous
  • anonymous
Hm, it seems much more simpler to understand if you apply those identities above, do you understand them @xoamb
anonymous
  • anonymous
\[(a+b)^2=a^2+b^2+2ab\] \[(a-b)^2=a^2+b^2-2ab\]
anonymous
  • anonymous
so it would be d?
anonymous
  • anonymous
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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