Please explain the proofs for sin(90 - θ)

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The discussion focuses on the proofs for the trigonometric identity sin(90 - θ) = cos(θ) as illustrated in the figures from "Theory and Problems of Plane and Spherical Trigonometry" by Frank Ayres. The user seeks clarification on how positive values for sin(90 - θ) are derived from the second, third, and fourth images, despite the negative coordinates in those figures. The key takeaway is that the identity holds true across different quadrants, with specific angles leading to positive or negative values based on their respective positions.

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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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