DimaMazin Posted Wednesday at 12:32 PM Author Posted Wednesday at 12:32 PM Can such problem be solved? Area A =area of the green segment. x=-cos/(1-2cos) y=sin/(1-2cos) sin/(1-2cos)=(p-6a)/2
Genady Posted Thursday at 03:04 PM Posted Thursday at 03:04 PM On 3/5/2025 at 8:32 AM, DimaMazin said: Can such problem be solved? Area A =area of the green segment. x=-cos/(1-2cos) y=sin/(1-2cos) sin/(1-2cos)=(p-6a)/2 I don't think your equation is correct.
DimaMazin Posted Friday at 02:52 AM Author Posted Friday at 02:52 AM 11 hours ago, Genady said: I don't think your equation is correct. Maybe you think sin and cos are sin(a) and cos(a)? No. sin is sin(p/2 +a) and cos is cos(p/2 +a) as it showed on diagram.
DimaMazin Posted Friday at 08:44 AM Author Posted Friday at 08:44 AM (edited) y = cos(a)/(1+2sin(a)) cos(a)/(1+2sin(a))=(p-6a)/2 Edited Friday at 08:49 AM by DimaMazin
Genady Posted Friday at 12:03 PM Posted Friday at 12:03 PM 3 hours ago, DimaMazin said: y = cos(a)/(1+2sin(a)) cos(a)/(1+2sin(a))=(p-6a)/2 My starting equation is \(\frac b 2 - \frac {\sin b \sin c} {2 \sin d} = \frac e 2 - \frac {\sin e} 2\) where \(b=2a\) \(2c=\pi - (\frac {\pi} 2 + a)\) \(d = \pi - b - c\) \(e = \frac {\pi} 2 - a\) I don't know if it's the same as yours.
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