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Please explain the proofs for sin(90 - θ)

  1. Sep 29, 2013 #1
    Please see the attached image. There are 6 figures.

    Please explain the proofs for sin(90 - θ) w.r.t the 2nd, 3rd and 4th images. I understand the proof w.r.t 1st image.

    In the 2nd image y1 and x is negative. In 3rd image y, y1, x, x1 are all negative and in the 4th image y and x1 are negative. How he gets +ve value for sin(90 - θ) w.r.t 2nd, 3rd and 4th figures?

    In the 3rd fig. -y1 = -x and -y = -x1.

    -y1/r1 = -x/r = -cos(θ)

    attachment.php?attachmentid=62294&d=1380475523.jpg
     

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  2. jcsd
  3. Sep 29, 2013 #2

    HallsofIvy

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    I don't understand your question. If, as in #2, the top figure on the right, [itex]90< \theta< 180[/itex], [itex]sin(90- \theta)[/itex] is certainly negative. Where does he say that it is positive?
     
  4. Sep 30, 2013 #3
    He is proving with the top four figures that sin(90 - θ ) = cos(θ). It is from the book "Theory and Problems of Plane and Spherical Trigonometry by Frank Ayres, Schaum's Outline Series"

    The book is available here. http://archive.org/details/SchaumsTheoryProblemsOfTrigonometry

    Please explain the proofs in page nos 56 and 57.

    In the 2nd figure from top ∠AOP = (180 - θ). How does he say ∠BOP1 = (90 - θ)?

    and ∠OAP and ∠OBP1 = 90.

    attachment.php?attachmentid=62317&d=1380518060.jpg
     

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    Last edited: Sep 30, 2013
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