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Trig identities

  1. Oct 22, 2004 #1
    I have basic trignometry doubt

    how to u get 5cos(2t-53.13degrees) from 3cos2t + 4 sin2t.

    Can someone suggest a site from where I can lean basic trinometry

    thanks.
    skan
     
  2. jcsd
  3. Oct 22, 2004 #2
    well, nice question...
    use cos(x-y)=cos(x)cos(y) + sin(x)sin(y).

    3cos2t + 4sin2t = 3(cos2t + (4/3)sin2t). The trick is to define 4/3 = tan(y)=siny/cosy where y = 53,13°. Now you have 3(cos2t + (siny/cosy)sin2t). or this becomes following expression : (3/cosy)(cosy * cos2t + siny * sin2t) =
    (3/cosy)cos(2t - 53.13°) and cosy = cos(53.13°) = 0.6

    3/cosy = 3/0.6 = 5

    problem solved

    regards
    marlon
     
  4. Oct 22, 2004 #3
  5. Oct 22, 2004 #4
    Thanks a lot for the answer and the great link!!!
     
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