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chaotic_butterflies

  • one year ago

(sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2 I've worked this out multiple times, and my answer isn't a possible choice...

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  1. chaotic_butterflies
    • one year ago
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    Answer choices: A) 1 B) 2 C) sin^2 theta D) cos^2 theta

  2. Nnesha
    • one year ago
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    show ur work please :-)

  3. Nnesha
    • one year ago
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    remember these two formulas \[\huge\rm (a+b)^2 = a^2 +2ab+b^2\] \[\huge\rm (a-b)^2 = a^2 -2ab+b^2\]

  4. chaotic_butterflies
    • one year ago
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    Oh goodness, I came at this a couple of different ways. At first I tried to replace sin and cos theta with regular numbers to see if the setup of the problem was some sort of rule, and that didn't come out right.

  5. Nnesha
    • one year ago
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    alright so let sin theta = a cos theta = b so \[\huge\rm (a-b)^2 =???\]

  6. chaotic_butterflies
    • one year ago
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    a^2 - b^2?

  7. Nnesha
    • one year ago
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    nope...

  8. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha remember these two formulas \[\huge\rm (a+b)^2 = a^2 +2ab+b^2\] \[\huge\rm (a-b)^2 = a^2 -2ab+b^2\] \(\color{blue}{\text{End of Quote}}\)

  9. Nnesha
    • one year ago
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    square of first number and square of 2nd term thne MULTIPLY BOTH TERMS by 2

  10. Nnesha
    • one year ago
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    or in other words (a-b)^2 is same as (a-b)(a-b) so you can foil :-)

  11. chaotic_butterflies
    • one year ago
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    Okay, I can at least do that...

  12. Nnesha
    • one year ago
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    ohh nice :-)

  13. Nnesha
    • one year ago
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    let me know what you get :-)

  14. chaotic_butterflies
    • one year ago
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    \[(\sin \theta - \cos \theta) (\sin \theta - \cos \theta) + (\sin \theta + \cos \theta) (\sin \theta + \cos \theta)\]

  15. chaotic_butterflies
    • one year ago
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    Well, that's at least the first step... am I wrong or doesn't that look like it should cancel out? @Nnesha

  16. Nnesha
    • one year ago
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    yes that's right now apply foil method :-)

  17. Nnesha
    • one year ago
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    no no you can't cancel anything yet

  18. Nnesha
    • one year ago
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    let sin = x cos = y it will be easy like simple algebra :-) (x-y)(x-y ) :-)

  19. chaotic_butterflies
    • one year ago
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    Oh phoey... well I'll continue to solve. If you don't mind, I like to simplify by grouping. (x-y)(x-y)+(x+y)(x+y) x^2 - xy (-y) (x-y) + (x+y)(x+y) x^2 - xy - yx + y^2 + (x+ y) (x+y) x^2 - xy - yx + y^2 + x^2 + xy (y)(x+y) x^2 - xy - yx + y^2 + x^2 + xy + yx + y^2 x^4 + y^2

  20. chaotic_butterflies
    • one year ago
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    *y^4

  21. Nnesha
    • one year ago
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    alright good job thanks! now x^2 +y^2 +x^2 +y^2 here COMBINE LIKE terms when you MULTIPLY same bases THEN you should add their exponent

  22. Nnesha
    • one year ago
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    \[\huge\rm x^2+y^2+x^2+y^2= ??\] combine like terms

  23. chaotic_butterflies
    • one year ago
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    Did I not already do that?

  24. Nnesha
    • one year ago
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    nope last step is wrong don't add their exponent here is a different COMBINE like terms : x+x = 2x Multiply x times x= x^2 there is a plus sign so you can't add their exponents

  25. chaotic_butterflies
    • one year ago
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    Goodness... I need to fresh up on my basic algebra >.<

  26. Nnesha
    • one year ago
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    alright so combine like terms meaning add coefficient of same bases

  27. chaotic_butterflies
    • one year ago
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    So 2x^4 + 2y^4..?

  28. Nnesha
    • one year ago
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    yeah but exponent should the same remember exponent rule \[\large\rm x \times x = x^2\] when u MULTIPLY same bases then u should add exponents

  29. chaotic_butterflies
    • one year ago
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    Oh wait it would just be 2x^2 + 2y^2 wouldn't it?

  30. chaotic_butterflies
    • one year ago
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    That loaded too late^

  31. Nnesha
    • one year ago
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    yes that's right

  32. chaotic_butterflies
    • one year ago
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    So that leaves me with \[\sin \theta ^{2} + \cos \theta ^{2}\]

  33. Nnesha
    • one year ago
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    now 2 is common factor s take it out \[\huge\rm 2x^2+2y^2\]\[\large\rm 2(x^2+y^2)\]

  34. Nnesha
    • one year ago
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    yes right change back to sin and cos so remember special identity sin^2 x + cos^2 x = what ?

  35. chaotic_butterflies
    • one year ago
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    Ooof... forgot about that!

  36. chaotic_butterflies
    • one year ago
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    I honestly don't remember.

  37. Nnesha
    • one year ago
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    hmm that's the only you shouldn't forget \[\huge\rm sin^2 \theta + \cos^2 \theta =1\]

  38. Nnesha
    • one year ago
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    so replace sin^2 + cos ^2 by 1 :-)

  39. chaotic_butterflies
    • one year ago
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    So it's just 2*1 = 2?

  40. Nnesha
    • one year ago
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    yes right :-) 2!!

  41. chaotic_butterflies
    • one year ago
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    Yay! Thank you for being so detailed, it really helped

  42. Nnesha
    • one year ago
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    my pleasure :-) gO_od job!! :=)

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