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Numerade Educator

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Problem 31 Hard Difficulty

Use the Mean Value Theorem to prove the inequality
$ | \sin a - \sin b | \leqslant | a - b | $ for all $ a $ and $ b $

Answer

Let $f(x)=\sin x$ and let $b<a$. Then $f(x)$ is continuous on $[b, a]$ and differentiable on $(b, a) .$ By the Mean Value Theorem,
there is a number $c \in(b, a)$ with $\sin a-\sin b=f(a)-f(b)=f^{\prime}(c)(a-b)=(\cos c)(a-b) .$ Thus,
$|\sin a-\sin b| \leq|\cos c||b-a| \leq|a-b| \cdot$ If $a<b,$ then $|\sin a-\sin b|=|\sin b-\sin a| \leq|b-a|=|a-b| .$ If $a=b,$ both
sides of the inequality are 0

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Video Transcript

So we're being asked to use the mean value item To prove the inequality, you have to sign a very minus sign of peace lesson equally after value A minus B for A and B since F. Beck, we can let f of x equal sign of X and we know that a derivative of effort Beck is close on X and we can actually, since signing close eye necks are for domestic, I mean, our continuous indefensible. We can actually apply them in various healing right away so we can write some C in the prime of eggs, which is co sign ex co sign of C. Their existence. See, just at the average slope is equal to the function was a sign of B minus sign of a over B minus A s O. We also know that the co sign just statement right here. So this statement, I hear co sign that that's him. But negative is bounded, which were negative one and a positive one and this statement canasta be rewritten in terms of ah, absolute value signed because the absolute value symbol actually represent distance. You can think of absolute values of distant and so this statement right here or the exact are the same. So we're actually going to rewrite the statement here, And since we applied the absent very function to this inside, we apply it to, um to the right side as well. So we have to We write this as, um, absolutely frozen, See is equal to the absolute value of sign of B miners. Sign of a all over baby might say. And now we're gonna go to the next page. Recall that the absolute coast I'm c'est les an equal one. So we're actually gonna substitute that one into our equation. So we're going to sign a minus. I mean, I'm so sorry, Sinan. Kind of b finest kind of a all over. Okay, B minus is less than equal to one. And then we're going to multiply, be minus a on both sides. We're going to sign a oh, my anti sign of B for that minus the sign of a And this is also still in absolute value. And then the B minus days also an absolute value. So I banned him. And then we can rewrite this as sign a minus a sign of the A minus B. And that is our proof. We have now proven our statement