Hello /sci/. I need some help, How do I isolate θ from this equation?

Use:
sin(u-v) = sin(u)cos(v) + sin(v)cos(u)
and
cos(u-v) = cos(u)cos(v) + sin(u)sin(v)
Sorry if no latex

so far ive managed to reduce the expression inside the cos^-1 to:
[math]\frac{1}{2}(\cos\theta(\cos\frac{\pi}{n}+1)+\cos\frac{4\pi}{n}-1)[/math]

Mathematica says it's

[eqn]
2 \left(\tan ^{-1}\left(\pm\frac{\sqrt{\cos \left(\frac{720}{n}\right)-\cos \left(\frac{180 \left(A+\pi n r^2-2 \pi r^2\right)}{\pi n r^2}\right)}}{\sqrt{\cos \left(\frac{720}{n}\right)+1}},\pm\frac{\sqrt{\cos \left(\frac{180 \left(A+\pi n r^2-2 \pi r^2\right)}{\pi n r^2}\right)+1}}{\sqrt{\cos \left(\frac{720}{n}\right)+1}}\right)+2 \pi c_1+90\right)
[/eqn]

>>8798203
sin x = (e^ix - e^-ix)/2i
cos x = (e^ix + e^-ix)/2

>>8798314
Thanks for the reply.
what does the c1 at the end of the equation mean?

>>8798351
Arbitrary integer constant. Adding integer multiples of [math]2 \pi[/math] inside of a trig function doesn't change the answer, so you can just pretend it's c_1 = 0, but other values can work.

>>8798314
i feel like this could still be simplified

I find working with radians is easier than degrees, so i will briefly change to radians
[eqn]\sin^2\left(\frac{\pi}{2}-\frac{\theta}{2}\right)\cos\frac{4\pi}{n}-\cos^2\left(\frac{\pi}{2}-\frac{\theta}{2}\right)=\left(\cos^2\frac{\theta}{2}\right)\left(\cos^2\frac{2\pi}{n}-\sin^2\frac{2\pi}{n}\right)-\left(\sin^2\frac{\theta}{2}\right)[/eqn]
[eqn]=\frac{1}{4}\left(\cos\left(\frac{\theta}{2}+\frac{2\pi}{n}\right)+\cos\left(\frac{\theta}{2}-\frac{2\pi}{n}\right)\right)^2-\frac{1}{4}\left(\sin\left(\frac{\theta}{2}+\frac{2\pi}{n}\right)-\sin\left(\frac{\theta}{2}-\frac{2\pi}{n}\right)\right)^2-\sin^2\frac{\theta}{2}[/eqn]
[eqn]=\frac{1}{4}\left(\cos\left(\theta+\frac{\pi}{n}\right)+\cos\left(\theta-\frac{\pi}{n}\right)+2\cos\frac{4\pi}{n}\right)+\frac{1}{2}(-1+\cos\theta)[/eqn]
[eqn]=\frac{1}{2}\left(\cos\theta\cos\frac{\pi}{n}+\cos\frac{4\pi}{n}-1+\cos\theta \right)[/eqn]

And that is equal to >>8798302

Now, rearranging the rest:

[eqn]\arccos(...)=180\left(\frac{A}{r^2}+(n-2)\pi\right)[/eqn]
[eqn]\Rightarrow \frac{1}{2}\left(\cos\theta\left(\cos\frac{180}{n}+1\right)+\cos\frac{720}{n}-1\right)=\cos\left(180\left(\frac{A}{r^2}+(n-2)\pi\right)\right)[/eqn]
And now you can easily move the terms
[eqn]\Rightarrow \cos\theta=\frac{2\cos\left(180\left(\frac{A}{r^2}+(n-2)\pi\right)\right)+1-\cos\frac{720}{n}}{\cos\frac{180}{n}+1}[/eqn]

if n is an integer this will be a lot easier

>>8798445
It describes a number of edges on a spherical polygon.

By the way. in one case:
n=6
θ=40,81
A=10,76

>>8798314
>atan2

>>8798203
lazy fuck do your homework yourself

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42KB, 500x500px

>>8798203
use a theta blocker