Question 1120989
Taylor's series: Given x is close to a,  {{{ f(x)=f(a)+f'(a)(x-a)+f''(a)(x-a)^2/2!+...+f^n(a)(x-a)^n/n!}}} 
Since you are estimating, take the first three above, and set a=pi/4, so x-a=-2 deg=-pi/90.
 {{{ sin(43deg)=sqrt(2)/2+sqrt(2)/2*(-pi/90)+(-sqrt(2)/2)*(-pi/90)^2/2}}}=0.6819933...
Approximately it will be happy to be 0.682. 
BTW the difference between this approximation and the exact value is contained within 5.06*10^-6.

And there're other fast ways to do that.
>Apply Maclaurin's series instead(take a=0). A lot faster.
>A much more simple method: sin(43deg)=sin(pi/4)-(sin(pi/4))'*(pi/90). Super fast and you'll be impressed by its accuracy.