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anonymous

  • one year ago

Simplify: (sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2

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  1. anonymous
    • one year ago
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    (sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2 (sin^2-cos^2) + (sin^2+cos^2) (y^2-x^2) + (y^2+x^2) eliminate x^2 (y^2+y^2)= 2sin^2

  2. anonymous
    • one year ago
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    please check my answer

  3. anonymous
    • one year ago
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    @Luigi0210 @KyanTheDoodle @mathmath333 @sleepyjess

  4. Nnesha
    • one year ago
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    \[\huge\rm (x+y)^2 \cancel= x^2 + y^2\]

  5. Nnesha
    • one year ago
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    nope let sin = x cos = y so \[(x -y)^2=(x-y)(x-y)\] now foil

  6. anonymous
    • one year ago
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    My options: A. −sin2 Θ B. −cos2 Θ C. 0 D. 2

  7. Nnesha
    • one year ago
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    alright so u didn't foil it right \[(x-y)^2 \cancel= x^2 -y^2\] (x-y)^2 is same as (x-y)(x-y ) can you foil these ?

  8. anonymous
    • one year ago
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    one moment so i can foil

  9. Nnesha
    • one year ago
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    yes right

  10. Pawanyadav
    • one year ago
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    Open the square Then put sin^2x+cos^x=1 You get 1-2sinxcosx +1+2sinxcosx =1+1=2 (I'm putting theta as x)

  11. anonymous
    • one year ago
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    is it x^2+2xy+y^2

  12. Pawanyadav
    • one year ago
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    Yes

  13. Nnesha
    • one year ago
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    yes right (x+y) = x^2 +2xy +y^2 :-)

  14. Nnesha
    • one year ago
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    what about (x-y)^2 ??

  15. anonymous
    • one year ago
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    the same thing

  16. anonymous
    • one year ago
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    (x-y)^2 is what i solved for

  17. Nnesha
    • one year ago
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    expand you have to foil it (X-y)^2 is same as (x-y)(x-y)

  18. Pawanyadav
    • one year ago
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    No 2xy is replaced by-2xy

  19. anonymous
    • one year ago
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    right

  20. Nnesha
    • one year ago
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    just like (x+y)(x+y)

  21. Nnesha
    • one year ago
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    \[\huge\rm x^2 -2xy +y^2 + x^2 +2xy +y^2\]\] combine like terms

  22. anonymous
    • one year ago
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    2x^2 + 2y^2

  23. Pawanyadav
    • one year ago
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    I hope you can add the two square carefully.

  24. Nnesha
    • one year ago
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    yep now replace x and y by sin and cos \[\huge\rm 2 \sin^2 + 2\cos^2\] take out common factor apply this identity \[\large\rm sin^2 \theta + \cos^2 \theta = 1 \]

  25. anonymous
    • one year ago
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    =2

  26. Nnesha
    • one year ago
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    yes right \[\huge\rm \color{reD}{ 2} \sin^2 +\color{reD}{ 2}\cos^2\] \[\huge\rm \color{reD}{ 2} (\sin^2 \theta +cos^2 \theta)\] \[\huge\rm \color{reD}{ 2} (1) =2\]

  27. anonymous
    • one year ago
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    Thank you so much for your help @Nnesha i really appreciate it.

  28. Nnesha
    • one year ago
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    my pleasure :-)

  29. anonymous
    • one year ago
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    <3

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