For what values of x is it true that |sin(x)-x| < 0.01?
I have spend 20 min on this,with no results. I went the Taylor expansion of sin(x), and it didn't help. When you get a solution, please post it. FYI, I graphed it, and was surprised. It reminded me how accurate the small angle approximation is.
I looked at the solution to
|sinx -x| = .01 and using Newton's Method found
x = ±.392493378
then testing, I found the above to be true for
-.3925 < x < + .3924
x is in radians
(for sinx - x = .01 my iteration expression was
(xcosx - sinx + .01)/(cosx - 1) )
To find the values of x that satisfy the inequality |sin(x) - x| < 0.01, we need to consider two cases: when sin(x) - x is positive, and when it is negative.
Case 1: sin(x) - x > 0
In this case, the inequality becomes sin(x) - x < 0.01.
To solve this inequality, we can add x to both sides:
sin(x) < x + 0.01.
Now, to determine the values of x that satisfy this inequality, we need to consider the range of sin(x) and determine where it is less than x + 0.01.
Since the range of sin(x) is [-1, 1], we can write the inequality as:
-1 < x + 0.01.
Subtracting 0.01 from both sides, we get:
-1 - 0.01 < x.
Simplifying further:
-1.01 < x.
Therefore, for the values of x where sin(x) - x > 0, the inequality is satisfied when x is greater than -1.01.
Case 2: sin(x) - x < 0
In this case, the inequality becomes -(sin(x) - x) < 0.01.
Multiplying both sides by -1, we get:
sin(x) - x > -0.01.
To determine the values of x that satisfy this inequality, we need to consider the range of sin(x) and determine where it is greater than -0.01.
Since the range of sin(x) is [-1, 1], we can write the inequality as:
x - 0.01 < sin(x).
Adding 0.01 to both sides, we get:
x < sin(x) + 0.01.
Therefore, for the values of x where sin(x) - x < 0, the inequality is satisfied when x is less than sin(x) + 0.01.
Combining the results from both cases, we find that the values of x satisfying the inequality |sin(x) - x| < 0.01 are:
x < sin(x) + 0.01, for x < -1.01
and
x > -1.01, for x > sin(x) + 0.01.