viii. use the mean value theorem to prove |sin b - sin a|≤|b - a|, for any a and b.
viii. use the mean value theorem to prove |sin b - sin a|≤|b - a|, for any a and b.
Answer
Explanation:
Step1: Define the function
Let (f(x)=\sin x). It is continuous on the closed - interval ([a,b]) (or ([b,a]) if (b < a)) and differentiable on the open - interval ((a,b)) (or ((b,a))).
Step2: Apply the Mean Value Theorem
By the Mean Value Theorem, there exists a (c\in(a,b)) (or ((b,a))) such that (f(b)-f(a)=f^{\prime}(c)(b - a)). Since (f(x)=\sin x), then (f^{\prime}(x)=\cos x), so (\sin b-\sin a=\cos c(b - a)).
Step3: Take the absolute - value
Take the absolute - value of both sides: (|\sin b-\sin a| = |\cos c|\cdot|b - a|).
Step4: Use the property of cosine function
We know that (|\cos c|\leq1) for all real numbers (c). So (|\sin b-\sin a|=|\cos c|\cdot|b - a|\leq|b - a|).
Answer:
The inequality (|\sin b-\sin a|\leq|b - a|) is proved.