Please explain the proofs for sin(90 - θ)

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The discussion focuses on understanding the proofs for the identity sin(90 - θ) using specific figures from a trigonometry book. The second, third, and fourth images illustrate scenarios where certain coordinates are negative, raising questions about how sin(90 - θ) can yield a positive value. It is clarified that sin(90 - θ) equals cos(θ), and the confusion stems from the interpretation of angles in different quadrants. The participants seek further explanation on specific angles and relationships depicted in the figures, particularly regarding the angles AOP and BOP1. The conversation emphasizes the need for clarity in geometric proofs related to trigonometric identities.
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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)

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I don't understand your question. If, as in #2, the top figure on the right, 90< \theta< 180, sin(90- \theta) is certainly negative. Where does he say that it is positive?
 
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.

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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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