HOW TO FIX PI
		
	
	
		09-19-2016, 11:56 PM 
	
	
		Let's say that for some reason, you are in a chat with a robot which has redefined all of math. For instance,
	
	
	
- pi = "fucking"
 
- e = 3
 
- sin = infinity
 
- cos = [3, 4.33, "meowing"]
 
- exp = ["e", "x", "p"]
 
- infinity = 2
 
- NaN = 10
 
- false = true
 
- OR JUST ANYTHING = "all star"
- pi=4*(.5!)^2
 
- sqrt(x)=x^.5
 
- i=sqrt(-1)- (OPTIONAL, if re isn't screwed up.) i=i-re(i)
 
- e=(i^i)^(-2/pi)
 
- exp(x)=e^x
 
- sin(x)=(exp(i*x)-exp(-i*x))/(2*i), cos(x)=.5*(exp(i*x)+exp(-i*x))
 
- tan(x)=sin(x)/cos(x), cot(x)=cos(x)/sin(x), sec(x)=1/cos(x), csc(x)=1/sin(x)- Assuming you have atan2 and it isn't screwed up (or you have arg, which is the same), you can fix all inverse trig
 
- atan(x)=atan2(x,1) OR atan(x)=arg(1+x*i)
 
- asin(x)=atan(x/sqrt(1-x^2)), acos(x)=atan(sqrt(1-x^2)/x)
 
- acot(x)=atan(1/x), asec(x)=acos(1/x), acsc(x)=asin(1/x)- If you have any inverse hyperbolic function or the natural logarithm, you can get all these functions back too:
 
- asinh(x)=acosh(sqrt(x^2+1)), acosh(x)=asinh(sqrt(x^2-1))
 
- atanh(x)=asinh(x/sqrt(1-x^2)) OR acosh(1/sqrt(1-x^2))
 
- asinh(x)=atanh(x/sqrt(1+x^2)), acosh(x)=atanh(sqrt(x^2-1)/x)
 
- log(x)=asinh((x^2-1)/(2*x)) OR acosh((x^2+1)/(2*x)) OR atanh((x^2-1)/(x^2+1))
 
- atanh(x)=.5*log((1+x)/(1-x)), acosh(x)=.5*log((x+1)/(x-1))


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