Okay, so let f(x)=sinx and let c be a point such that a<c<b.

Apply the mean value theorem.

Knowing that the function cos(x) is always decreasing, what can you say about cos(a), cos(c), and cos(b)?

Reply 4

sahil112

OP

13

so would we have something like this to be working with:

sinb-sina/b-a goes to cosb-cosa/b-a? and then work from there

Reply 5

aznkid66

1

Um...not exactly.

The mean value theorem states that in the bounds from a to b there exists a c such that a<c<b and f'(c)=[f(b)-f(a)]/[b-a].

So let f(x)=sin(x).

According to the mean value theorem, between the a and b values given in the problem, there exists a c such that:

cos(c)=[sin(b)-sin(a)]/[b-a]

Now, what do you know about cos(a), cos(b), and cos(c) given that the function is always decreasing from a to b and a<c<b?

Reply 6

sahil112

OP

13

cosc is going to lie between cosa and cosb?

Reply 7

aznkid66

1

Yup, but more specifically you know the direction of the inequality.

So now you have:

cos(b)<cos(c)<cos(a)

And:

cos(c)=[sin(b)-sin(a)]/[b-a]

So clearly:

cos(b)<[sin(b)-sin(a)]/[b-a]<cos(a) QED

Do you see what sub in for a and b to do the next part?

Reply 8

sahil112

OP

13

Original post by aznkid66

Yup, but more specifically you know the direction of the inequality.

So now you have:

cos(b)<cos(c)<cos(a)

And:

cos(c)=[sin(b)-sin(a)]/[b-a]

So clearly:

cos(b)<[sin(b)-sin(a)]/[b-a]<cos(a) QED

Do you see what sub in for a and b to do the next part?

i was confused about what the question was asking, so all it wanted was for us to just state that it comes back to what it first stated but by using the MVT?

um i was thinking pi/4 and pi/2 or isit asking us to sub in a function?

Reply 9

sahil112

OP

13

thank you for the help by the way

Reply 10

sahil112

OP

13

oh i got it, make b=2x and a=x

Reply 11

aznkid66

1

Well, yes, we were supposed to arrive "back to what it first stated" because we were asked to prove it to be true, not assume it to be true.

Also, I was thinking of different values for a and b. Are you sure the ones you suggested work? ._.

It's quite obviously that a substitution where cosb=cosx, b=...? and cosa=1, a=...? works as a specific case where the statement we just proved implies what we want to deduce.

Reply 12

sahil112

OP

13

Original post by aznkid66

Well, yes, we were supposed to arrive "back to what it first stated" because we were asked to prove it to be true, not assume it to be true.

Also, I was thinking of different values for a and b. Are you sure the ones you suggested work? ._.

It's quite obviously that a substitution where cosb=cosx, b=...? and cosa=1, a=...? works as a specific case where the statement we just proved implies what we want to deduce.

ah sorry jumped the gun, yes b=x and a=1 would work :smile:

thank you

Reply 13

sahil112

OP

13

my bad a=0 i mean

mean value theorem

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