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GucciDoTheDishes

  • one year ago

**medal and fan Simplify: (sin Θ − cos Θ)2 + (sin Θ + cos Θ)2

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  1. guccidothedishes
    • one year ago
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    answer choices A) 1 B) 2 C)sin^2Θ D)cos^2

  2. Nnesha
    • one year ago
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    let sin theta = x cos theta = y so you have \[\huge\rm (x-y)^2 +(x +y)^2\]

  3. Nnesha
    • one year ago
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    (x-y)^2 is same as (x-y)(x-y) apply foil method

  4. guccidothedishes
    • one year ago
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    so it cancels out ?

  5. Nnesha
    • one year ago
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    no there is a plus sign

  6. guccidothedishes
    • one year ago
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    im not really familiar with the foil method

  7. guccidothedishes
    • one year ago
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    do I just multiply (x-y)(x-y) ??

  8. Nnesha
    • one year ago
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    ohh okay |dw:1438028749837:dw| multiply first and 2nd term of first parentheses by 2nd parentheses

  9. Nnesha
    • one year ago
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    yees right

  10. guccidothedishes
    • one year ago
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    I know part of it is X^2 so x &y multiplied its just x*y ?

  11. Nnesha
    • one year ago
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    yes right but don't forget the negative sign \[\huge\rm (x-y)(x-y)\] |dw:1438029051942:dw|

  12. Nnesha
    • one year ago
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    x times -y = -xy

  13. guccidothedishes
    • one year ago
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    okay but now whats the next step to the problem ?

  14. Nnesha
    • one year ago
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    now multiply 2nd parentheses by 2nd term of 1st parentheses|dw:1438029177493:dw|

  15. guccidothedishes
    • one year ago
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    -yx - y^2

  16. Nnesha
    • one year ago
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    yes right

  17. Nnesha
    • one year ago
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    well no little mistake -y times -y = ?

  18. guccidothedishes
    • one year ago
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    positive y ?

  19. Nnesha
    • one year ago
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    positive y^2

  20. guccidothedishes
    • one year ago
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    oh sorry

  21. Nnesha
    • one year ago
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    \[\huge\rm\color{reD}{ (x-y)^2} +(x +y)^2\] \[\huge\rm \color{Red}{x^2-xy-xy+y^2} +(x+y)^2\] combine like terms and then foil (x+y)^2

  22. guccidothedishes
    • one year ago
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    -xy and -xy are like terms do they cancel out ?

  23. Nnesha
    • one year ago
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    no you need to be careful abt signs -xy -xy you would cancel out if there are opposite signs -xy -xy = ? add example -1-1=-2

  24. guccidothedishes
    • one year ago
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    but I would add them together correct ?

  25. Nnesha
    • one year ago
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    yep

  26. guccidothedishes
    • one year ago
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    So i would get -2xy ?

  27. Nnesha
    • one year ago
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    yep right now multiply (x+y)^2

  28. Nnesha
    • one year ago
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    (x+y)^2 is same as (X+y)(x+y)

  29. guccidothedishes
    • one year ago
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    okay so now i ditribute x-y^2 to x^2 + (x+y)^2+y^2?

  30. freckles
    • one year ago
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    Have you expanded (x+y)^2 ? \[\huge\rm \color{Red}{x^2-xy-xy+y^2} +(x+y)^2\]

  31. guccidothedishes
    • one year ago
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    no ?

  32. freckles
    • one year ago
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    (x+y)^2 is the same as (x+y)(x+y) multiply this out

  33. freckles
    • one year ago
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    x(x+y)+y(x+y) use distributive property twice

  34. guccidothedishes
    • one year ago
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    so Im going to distribute the x to whats in parentheses and then do the same with the y , correct ?

  35. freckles
    • one year ago
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    yes that is the distributive property it says a(b+c)=ab+ac

  36. guccidothedishes
    • one year ago
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    okay after distributing I got x^2+xy + yx+y^2

  37. freckles
    • one year ago
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    yes and xy+xy=?

  38. guccidothedishes
    • one year ago
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    xy^2 ?

  39. freckles
    • one year ago
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    if have an apple and find another apple do you have a square amount of apples or just 2 apples?

  40. guccidothedishes
    • one year ago
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    so 2xy ?

  41. freckles
    • one year ago
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    yes

  42. freckles
    • one year ago
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    (x+y)^2=x^2+2xy+y^2 and (x-y)^2=x^2-2xy+y^2

  43. freckles
    • one year ago
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    also recall sin^2(u)+cos^2(u)=1

  44. freckles
    • one year ago
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    anyways let us know if you need further help on this one

  45. guccidothedishes
    • one year ago
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    but I still dont get it lol sorry , what would the next step be ?

  46. freckles
    • one year ago
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    recall you started with \[(\sin(\theta)+\cos(\theta))^2+(\sin(\theta)-\cos(\theta))^2 \\ \text{ then you expanded} \\ \sin^2(\theta)+2\sin(\theta)\cos(\theta)+\cos^2(\theta)+\sin^2(\theta)-2\sin(\theta)\cos(\theta)+\cos^2(\theta)\] are you not sure what sin^2(theta)+cos^2(theta) equals ? or is it you don't know what 2sin(theta)cos(theta)-2sin(theta)cos(theta) is?

  47. guccidothedishes
    • one year ago
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    im not sure what it equals

  48. freckles
    • one year ago
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    which one?

  49. freckles
    • one year ago
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    sin^2(theta)+cos^2(theta)? or the same thing - the same thing?

  50. freckles
    • one year ago
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    the first one is a Pythagorean identity sin^2(u)+cos^2(u)=1 the second one is using the additive inverse property that is u+(-u)=0 or this could be written as u-u=0 which should make since because if you have exactly one apple and so one takes that apple away you have no apples aka 0 apples

  51. guccidothedishes
    • one year ago
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    okay thank you

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