For approximately what values of x can you replace sin x by

Solution

Taylors theorem with Lagrange-error term for f(x) = sin(x) up to order 5 around a=0 is f(x) = x - x^3/6 + f^(5)(t)*x^5/120 where f^(5) is the 5th derivative of f(x), thus f^(5)(x) = cos(x) and t is some point in the interval (0,x) (if x>0) or (x,0) (if x<0). The magnitude of the error is |f^(5)(t)*x^5/120| = |cos(t)|*|x|^5/120. Now you don\'t know at which point t you will get the exact error, but you know that |cos(t)| <= 1 for all t, so by using that you can know for sure that the magnitude of the error is no greater than |x|^5/120. Now solving for x in |x|^5/120 = 0.0005 gives |x|^5 = 0.06 which is |x| = (0.06)^(1/5) = 0.569679... Now to make sure we don\'t exceed the error we have to round down, thus -0.56 < x < 0.56. But a equally good answer would have been -0.5696 < x < 0.5696 etc