Furthering tmonteil's answer

N will try to refine its answer until the `digits' digits no longer change after refinement, and prints the last refined value. The fact that it stops and returns an absolute value of the order of $1O^{-202}$ means that this value has $200$ exact digits, which are all 0. You cane therefore state confidently that this absolute value is inferior to $10^{-200}$, therefore not distingushable of $0$ with 200 digits precision.

Computing an exact value

You seem to aim at computing $\sin\frac{50\pi}{180}=\sin\frac{5\pi}{18}$. Given $\sin{}x$ and $\cos{}x$, `Sage' can easily compute $\sin{5x}$ :

sage: sin(5*x)==sin(5*x).trig_expand()
sin(5*x) == 5*cos(x)^4*sin(x) - 10*cos(x)^2*sin(x)^3 + sin(x)^5

Your problem is now therefore reduced to computing $\sin\frac{\pi}{18}$ and $\cos\frac{\pi}{18}$. Remarking that $\sin\frac{\pi}{2}=1\ \cos\frac{\pi}{2}=0$, we can compute them by noting that these quantities are roots of the polynomial equations :

Sys=[u(9*x).trig_expand()==u(pi/2) for u in (sin, cos)]
Sys
[9*cos(x)^8*sin(x) - 84*cos(x)^6*sin(x)^3 + 126*cos(x)^4*sin(x)^5 - 36*cos(x)^2*sin(x)^7 + sin(x)^9 == 1,
  cos(x)^9 - 36*cos(x)^7*sin(x)^2 + 126*cos(x)^5*sin(x)^4 - 84*cos(x)^3*sin(x)^6 + 9*cos(x)*sin(x)^8 == 0]

Here, we could try to solve this polynomial system in SR by typing solve([u.subs({sin(x):S, cos(x):C}) for u in Sys],[S,C]) ; this would give us a whole lot of solutions in $\mathbb{C}$, expressed using floating point numbers (i. e. inexact solutions).

A better solution is to get exact roots by moving to the ring of polynomials over the (real) algebraic number field. Sage is able to compute "exact" roots of such polynomials :

R1.<s,c>=AA[]
PSys=[R1((u.lhs()-u.rhs()).subs({sin(x):s, cos(x):c})) for u in Sys]
PSys

[s^9 - 36*s^7*c^2 + 126*s^5*c^4 - 84*s^3*c^6 + 9*s*c^8 - 1,
  9*s^8*c - 84*s^6*c^3 + 126*s^4*c^5 - 36*s^2*c^7 + c^9]

The roots of this polynomial system are given by :

PSol=R1.ideal(PSys).variety()

Note that there are 9 such real solutions. To select the right one, we can use a tad of elementary geometric intuition and :

• restrict ourselves to the first quadrant (i. e. s' andc' both positive) ;

• note that in this quadrant, pi/18 has the smallest sine and the largest cosine.

It turns out that both have a radical expression :

Q1=[u for u in PSol if u[s]>0 and u[c]>0]
S1=min([u[s] for u in Q1])
C1=max([u[c] for u in Q1])
(S1.radical_expression(), C1.radical_expression())

(-1/2*(I*sqrt(3) + 1)*(1/16*I*sqrt(3) - 1/16)^(1/3) - 1/8*(-I*sqrt(3) + 1)/(1/16*I*sqrt(3) - 1/16)^(1/3),
 1/4*(4*(1/128*I*sqrt(3) + 1/128)^(1/3) + 1)/(1/128*I*sqrt(3) + 1/128)^(1/6))

Computing $\sin\frac{5\pi}{18}$ is now trivial :

S50=sin(5*x).trig_expand().subs({sin(x):S1, cos(x):C1})
S50

0.7660444431189781?

Note that this is not a mere numerical approximation, but a representation of an algorithm allowing to compute numerical approximations to arbitrary precision, thus representing a unique algebraic real number.

This result can be exactly verified:

bool(S50==sin(50*pi/180))

True

HTH,

I can not reproduce your computation (the definition of s seems uncomplete), but you should notice that -8e-202 is extremly small (in particular, the first 200 digits of s and N(sin(50/180*pi)-s,digits=200) are equal).

If you want a result with an explicit bound on the error, you should probably use RealBallField or RealIntervalField.