Use Mean Value Theorem to show |sin(a)-sin(b)|<=|a-b|

Limits, differentiation, related rates, integration, trig integrals, etc.
nona.m.nona
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Use Mean Value Theorem to show |sin(a)-sin(b)|<=|a-b|

Postby nona.m.nona » Mon Jan 26, 2009 1:16 pm

Use the Mean Value Theorem to prove the inequality |sin(a) - sin(b)| <= |a - b| for all a and b.

The MVT says that there is some c between a and b so sin(a) - sin(b) = cos(c)(a - b). Where do I go from there? :confused:

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stapel_eliz
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Postby stapel_eliz » Mon Jan 26, 2009 1:33 pm

nona.m.nona wrote:Use the Mean Value Theorem to prove the inequality |sin(a) - sin(b)| <= |a - b| for all a and b.

The MVT says that there is some c between a and b so sin(a) - sin(b) = cos(c)(a - b).

If the two sides are equal, then their absolute values are certainly equal:

. . . . .sin(a) - sin(b) = cos(c)(a - b)

. . . . .|sin(a) - sin(b)| = |cos(c)(a - b)|

You can then split the absolute value on the right-hand side:

. . . . .|sin(a) - sin(b)| = |cos(c)||a - b|

You would like to have the following:

. . . . .|sin(a) - sin(b)| <= |a - b|

Think about the cosine function, and the range of values it takes on. (Hint: |a - b| = (1)|a - b|.) :wink:

Eliz.


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