Half-Angle Formulas...4

Discussion in 'Geometry and Trigonometry' started by nycmathguy, Jan 2, 2022.

  1. nycmathguy

    nycmathguy

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    Section 5.5

    Screenshot_20211231-174449_Samsung Notes.jpg

    IMG_20220101_202731.jpg

    IMG_20220101_202743.jpg

    IMG_20220101_202752.jpg
     
    nycmathguy, Jan 2, 2022
    #1
  2. nycmathguy

    MathLover1

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    sin(u)=5/13-> opp=5, hyp=13=> adj=sqrt(13^2-5^2)=12
    cos(u)=12/13

    sin(u/2)=sqrt((1-cos(u))/2)
    sin(u/2)=sqrt((1-12/13)/2)
    sin(u/2)=sqrt((1/13)/2)
    sin(u/2)=sqrt((1/26)
    sin(u/2)=sqrt(26)/26
     
    MathLover1, Jan 2, 2022
    #2
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  3. nycmathguy

    nycmathguy

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    Where did I go wrong?
    Can you point out my error?
     
    nycmathguy, Jan 2, 2022
    #3
  4. nycmathguy

    MathLover1

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    while doing sin(u/2), you got a line

    =sqrt(1/26) which is correct, then your next line is

    =sqrt(1)/sqrt(13) instead sqrt(1)/sqrt(26)
     
    MathLover1, Jan 2, 2022
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  5. nycmathguy

    nycmathguy

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    Making errors leads to learning.
     
    nycmathguy, Jan 2, 2022
    #5
  6. nycmathguy

    nycmathguy

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    IMG_20220103_205014.jpg

    IMG_20220103_205023.jpg
     
    nycmathguy, Jan 4, 2022
    #6
  7. nycmathguy

    MathLover1

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    correct
     
    MathLover1, Jan 4, 2022
    #7
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  8. nycmathguy

    nycmathguy

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    Finally got it right.
     
    nycmathguy, Jan 4, 2022
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