#"This question is ill-posed as"#
#0.93969#
#"is a polynomial of degree 0 which does the job."#
#"A calculator calculates the value of cos(x) through the Taylor"#
#"series."#
#"The Taylor series of cos(x) is : "#
#1 - x^2/(2!) + x^4/(4!) - x^6/(6!) + ...#
#"What you need to know is that the angle you fill in this series"#
#"must be in radians. So 20° = "pi/9 = 0.349..." rad."#
#"To have a fast convergent series |x| must be smaller than 1,"#
#"by preference smaller than 0.5 even."#
#"We have luck as this is the case. In the other case we would"#
#"have to use goniometric identities to make the value smaller."#
#"We must have :"#
#(pi/9)^n/(n!) < 0.001 ", n as small as possible"#
#=> n=4#
#"This is the fault term so, "x^4/(4!)" does not have to be"#
#"evaluated even, so we need only the first two terms :"#
#1 - x^2/2 = 1 - (pi/9)^2/2 = 0.93908#
#"Clearly, the error is less than "10^-3" or "0.001"."#
#"You might ask yourself further how we get the value of "pi"."#
#"This can be done, among others, through the Taylor series of"#
#"arctan(x) as arctan(1) = "pi/4 => pi = 4*arctan(1)"."#
#"But there are other quicker (better convergent) series to"#
#"calculate "pi"."#
