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anonymous

  • one year ago

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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  1. anonymous
    • one year ago
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    \[(\sin(\theta)-\cos(\theta))^2-(\sin(\theta)+\cos(\theta))^2\] Apply the identity \[a^2-b^2=(a+b)(a-b)\] \[\implies ((\sin(\theta)-\cos(\theta))+(\sin(\theta)+\cos(\theta)))\times((\sin(\theta)-\cos(\theta))-(\sin(\theta)+\cos(\theta)))\]

  2. anonymous
    • one year ago
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    Here, we've taken \[a=\sin(\theta)-\cos(\theta)\] and \[b=\sin(\theta)+\cos(\theta)\]

  3. anonymous
    • one year ago
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    Wouldnt the answer be 0 tho? since it's cancelled out?

  4. anonymous
    • one year ago
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    Nope, open the parenthesis and find out for yourself

  5. anonymous
    • one year ago
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    i still dont understand

  6. Michele_Laino
    • one year ago
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    hint: you can use these identities: \[\Large \begin{gathered} {\left( {\sin \theta - \cos \theta } \right)^2} = {\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} - 2\sin \theta \cos \theta \hfill \\ \hfill \\ {\left( {\sin \theta + \cos \theta } \right)^2} = {\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} + 2\sin \theta \cos \theta \hfill \\ \end{gathered} \]

  7. anonymous
    • one year ago
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    Hm, it seems much more simpler to understand if you apply those identities above, do you understand them @xoamb

  8. anonymous
    • one year ago
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    \[(a+b)^2=a^2+b^2+2ab\] \[(a-b)^2=a^2+b^2-2ab\]

  9. anonymous
    • one year ago
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    so it would be d?

  10. anonymous
    • one year ago
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    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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