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Please explain the proofs for sin(90 - θ)

  • Thread starter jayanthd
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  • #1
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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
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?
 
  • #3
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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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