- Thread starter EvanJ
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theta = 2arcsin (c/2R)

c = 1.5 and R = 1, so C/2R = 0.75

arcsin(0.75) = about 48.59 degrees

theta = 2*48.59 = about 97.18 degrees

A = (1^2)/2 (97.18pi/180) - sin(97.18pi/180))

97.18pi/180 = about 1.696

A = 1/2 (1.696 - sin 1.696)

sin 1.696 = about 0.296

A = 1/2 (1.696 - 0.296) = 1/2 (1.666) = 0.833

I prefer degrees to radians, but I feel like I'm converting to radians and then taking sin of 1.696, which is wrong. However, if I change that part to sin of 97.18 degrees, I get an answer that's way too low:

A = (1^2)/2 (97.18pi/180) - sin(97.18pi/180))

97.18pi/180 = about 1.696

A = 1/2 (1.696 - sin 97.18)

sin 97.18 = about 0.992

A = 1/2 (1.696 - 0.992) = 1/2 (0.704) = 0.352

You possibly tried to find sin(1.696°), but that's 0.0296 approximately, not 0.296. Both are incorrect.sin 1.696 = about 0.296

You had already found that $\theta$ = about 97.18°, and sine of that is about 0.992, as you later found.

Alternatively, sine of 1.696 radians gives the same value.

Your calculation of 0.352 is correct.

from the diagram on the proof page, $\cos\left(\dfrac{\theta}{2}\right) = \dfrac{d}{R}$

double angle identity ... $\cos{\theta} = 2\cos^2\left(\dfrac{\theta}{2}\right) - 1$

$2R\sqrt{2-2\cos{\theta}} = 2R\sqrt{2 - 2\left[2\cos^2\left(\dfrac{\theta}{2}\right) - 1\right]} = 2R\sqrt{2 - 4\cos^2\left(\dfrac{\theta}{2}\right) + 2} = 2R\sqrt{4 - 4\cos^2\left(\dfrac{\theta}{2}\right)} = 2R\sqrt{4\left[1 - \cos^2\left(\dfrac{\theta}{2}\right)\right]} = 4R\sqrt{1 - \left(\dfrac{d}{R}\right)^2}$

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