Use the MVT to prove:
absolute value of (sin(a)-sin(b)) is less than or equal to the absolute value of (a-b) for all a,b
To prove the inequality using the Mean Value Theorem (MVT), we need to find a function that satisfies the conditions of the theorem.
Let's define a function, f(x) = sin(x). We want to find the derivative of this function in order to apply the MVT. The derivative of sin(x) is cos(x).
Now, let's consider two points, a and b, where a > b. According to the Mean Value Theorem, there exists a point c in the interval (b, a) such that the derivative of f at c is equal to the slope of the line connecting the points (a, f(a)) and (b, f(b)).
The slope of the line passing through these two points is given by (f(a) - f(b))/(a - b), which in this case simplifies to (sin(a) - sin(b))/(a - b).
Now, let's find the value of c. According to the MVT, we have:
cos(c) = (sin(a) - sin(b))/(a - b)
Since the absolute value of cos(c) is always less than or equal to 1, we can write:
|(sin(a) - sin(b))/(a - b)| <= 1
Multiplying both sides of the equation by |a - b|, we get:
|sin(a) - sin(b)| <= |a - b|
Hence, we have proven that the absolute value of (sin(a) - sin(b)) is less than or equal to the absolute value of (a - b) for all a, b.