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anonymous

  • one year ago

Simplify: (sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2. Options include: -sin^2 Θ -cos^2Θ 0 2

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  1. anonymous
    • one year ago
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    So what would you get should you foil out both of those?

  2. anonymous
    • one year ago
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    Ok I'm going to let sin Θ = x and cos Θ = y (x-y)(x-y) (x^2-xy+y^2) (x+y)(x+y) (x^2 +xy+y^2)

  3. anonymous
    • one year ago
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    so (sin Θ^2 - sin Θ times cos Θ + cos Θ^2) + (sin Θ^2 + sin Θ times cos Θ + cos Θ^2) but what is sin theta times cos theta??

  4. anonymous
    • one year ago
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    Well, you should have a \(-2\sin(\theta)\cos(\theta)\) and a \(+2\sin(\theta)\cos(\theta)\), so those would cancel anyway, we don't need to worry about them.

  5. anonymous
    • one year ago
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    ohh I forgot the outside values

  6. anonymous
    • one year ago
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    But once you cancel and combine everything, you would have \(2\sin^{2}(\theta) + 2\cos^{2}(\theta) = 2(\sin^{2}(\theta) + \cos^{2}(\theta))\)

  7. anonymous
    • one year ago
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    Okay so it should be (sin^2 theta + cos^2 theta) + (sin^2 theta + cos^2 theta)? And the pythagorean identity says that sin^2 theta + cos^2 theta = 1, so 1 + 1 = 2?

  8. anonymous
    • one year ago
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    Yep :)

  9. anonymous
    • one year ago
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    Thank you so much!

  10. anonymous
    • one year ago
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    Yep, no problem :3

  11. xapproachesinfinity
    • one year ago
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    i have seen this question here over and over lol did teacher run out of problems or something hehe

  12. xapproachesinfinity
    • one year ago
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    teachers*

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