Question

show that for any real numbers a and b,the inequality |sinb-sin a|<=|b-a| is true

i will wait for your help

Answer 1

You can use the Mean Value Theorem to prove this. Have you covered the MVT yet?

Answer 2

i know i should use it but how? can you explain me i need a proof

Answer 3

GIven a continous function f on a closed interval [a,b], which is also differentiable on the open interval (a,b), there exists a point c in the open interval (a,b) such that:\[f(b)-f(a)=f^\prime(c)(b-a)\]Thats what the mean value theorem states. The function for your question is sin(x). Try to see how you would apply the MVT to get what you want.

Answer 4

sinb-sina=cosc.(b-a)

Answer 5

thats right. If you apply absolute values to both sides of that equation we get:\[|\sin(b)-\sin(a)|=|\cos(c)|\cdot|b-a|\]now look at the "cos(c)" part. What can you say about the function cosine? is it bounded/unbounded? what is its max/min?

Answer 6

1,0 for max and min

Answer 7

right. So what you are saying is:\[|\cos(c)|\le 1\] for any real number c.

Answer 8

yes that's it ?

Answer 9

yep, thats pretty much it. Since cos(c) <=1, we see that:\[|\sin(b)-\sin(a)|=|\cos(c)||b-a|\le 1\cdot |b-a|=|b-a|\] This produces the desired result.

Answer 10

sorry i cant understand the last one

Answer 11

Since 1 is the biggest that cos(c) can be, if I substitute the cos(c) for a 1, the result is always bigger. Thats what this statement means:\[|\cos(c)||b-a|\le 1\cdot |b-a|\]

Answer 12

oh it's clearer right now :) thank you so much