Using the Mean Value Theorem, prove this inequality.

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Using the Mean Value Theorem, prove that

$|cos^2(b)-cos^2(a)|gt frac14|b-a| $

for all $a,b in (fracpi4,fracpi3)$

So far, I have

Let $f(x)=cos^2(x)$

Then $f(x)$ is continuous and differentiable on all $xin mathbbR$ so it is continuous and differentiable on $xin (fracpi4,fracpi3)$.

Applying the Mean Value Theorem to $f(x)$, we have:

$fraccos^2(b)-cos^2(a)b-a= -2cos(x)sin(x)$

$fraccos^2(b)-cos^2(a)b-a= -sin(2x)$

$|cos^2(b)-cos^2(a)| = -sin(2x)|b-a|$

Now I do not know how to go from here, as on the interval $(fracpi4,fracpi3)$, $-fracsqrt32lt-sin(2x)lt -1$.

Or am I making a mistake?

edited Aug 20 at 13:41

Bernard

111k635103

asked Aug 20 at 12:55

user499701

947

add a commentÂ |Â

up vote
1
down vote

favorite

Using the Mean Value Theorem, prove that

$|cos^2(b)-cos^2(a)|gt frac14|b-a| $

for all $a,b in (fracpi4,fracpi3)$

So far, I have

Let $f(x)=cos^2(x)$

Then $f(x)$ is continuous and differentiable on all $xin mathbbR$ so it is continuous and differentiable on $xin (fracpi4,fracpi3)$.

Applying the Mean Value Theorem to $f(x)$, we have:

$fraccos^2(b)-cos^2(a)b-a= -2cos(x)sin(x)$

$fraccos^2(b)-cos^2(a)b-a= -sin(2x)$

$|cos^2(b)-cos^2(a)| = -sin(2x)|b-a|$

Now I do not know how to go from here, as on the interval $(fracpi4,fracpi3)$, $-fracsqrt32lt-sin(2x)lt -1$.

Or am I making a mistake?

edited Aug 20 at 13:41

Bernard

111k635103

asked Aug 20 at 12:55

user499701

947

add a commentÂ |Â

up vote
1
down vote

favorite

Using the Mean Value Theorem, prove that

$|cos^2(b)-cos^2(a)|gt frac14|b-a| $

for all $a,b in (fracpi4,fracpi3)$

So far, I have

Let $f(x)=cos^2(x)$

Then $f(x)$ is continuous and differentiable on all $xin mathbbR$ so it is continuous and differentiable on $xin (fracpi4,fracpi3)$.

Applying the Mean Value Theorem to $f(x)$, we have:

$fraccos^2(b)-cos^2(a)b-a= -2cos(x)sin(x)$

$fraccos^2(b)-cos^2(a)b-a= -sin(2x)$

$|cos^2(b)-cos^2(a)| = -sin(2x)|b-a|$

Now I do not know how to go from here, as on the interval $(fracpi4,fracpi3)$, $-fracsqrt32lt-sin(2x)lt -1$.

Or am I making a mistake?

edited Aug 20 at 13:41

Bernard

111k635103

asked Aug 20 at 12:55

user499701

947

Using the Mean Value Theorem, prove that

$|cos^2(b)-cos^2(a)|gt frac14|b-a| $

for all $a,b in (fracpi4,fracpi3)$

So far, I have

Let $f(x)=cos^2(x)$

Then $f(x)$ is continuous and differentiable on all $xin mathbbR$ so it is continuous and differentiable on $xin (fracpi4,fracpi3)$.

Applying the Mean Value Theorem to $f(x)$, we have:

$fraccos^2(b)-cos^2(a)b-a= -2cos(x)sin(x)$

$fraccos^2(b)-cos^2(a)b-a= -sin(2x)$

$|cos^2(b)-cos^2(a)| = -sin(2x)|b-a|$

Now I do not know how to go from here, as on the interval $(fracpi4,fracpi3)$, $-fracsqrt32lt-sin(2x)lt -1$.

Or am I making a mistake?

edited Aug 20 at 13:41

Bernard

111k635103

asked Aug 20 at 12:55

user499701

947

edited Aug 20 at 13:41

Bernard

111k635103

edited Aug 20 at 13:41

Bernard

111k635103

edited Aug 20 at 13:41

Bernard

111k635103

asked Aug 20 at 12:55

user499701

947

asked Aug 20 at 12:55

user499701

947

asked Aug 20 at 12:55

user499701

947

add a commentÂ |Â

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edited Aug 20 at 13:13

answered Aug 20 at 13:07

Dr. Sonnhard Graubner

67.5k32660

add a commentÂ |Â

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edited Aug 20 at 13:13

answered Aug 20 at 13:07

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up vote
2
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edited Aug 20 at 13:13

answered Aug 20 at 13:07

Dr. Sonnhard Graubner

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up vote
2
down vote

edited Aug 20 at 13:13

answered Aug 20 at 13:07

Dr. Sonnhard Graubner

67.5k32660

edited Aug 20 at 13:13

answered Aug 20 at 13:07

Dr. Sonnhard Graubner

67.5k32660

edited Aug 20 at 13:13

answered Aug 20 at 13:07

Dr. Sonnhard Graubner

67.5k32660

answered Aug 20 at 13:07

Dr. Sonnhard Graubner

67.5k32660

answered Aug 20 at 13:07

Dr. Sonnhard Graubner

67.5k32660

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