Absolute value of sin(x)

Assume a>0 and b>0.

If a<r and b<s, then ab<sr. Now put what Dr Peterson gave you in his last post together.

Edit: | q*s | = |q| |s|
 
Are you asking whether it is true that [MATH]|xy| = |x|\cdot |y|[/MATH]? or whether, if [MATH]|x|\le a[/MATH] and [MATH]|y|\le b[/MATH], then [MATH]|xy|\le ab[/MATH]?

You can certainly multiply things, and various properties of absolute value and of inequality may apply ...

Give it a try! Often if you write something out, you can decide whether it is true. That works much better than sitting around waiting to be sure what you should do next.
 
Dr.Peterson said:
Are you asking whether it is true that [MATH]|xy| = |x|\cdot |y|[/MATH]? or whether, if [MATH]|x|\le a[/MATH] and [MATH]|y|\le b[/MATH], then [MATH]|xy|\le ab[/MATH]?

You can certainly multiply things, and various properties of absolute value and of inequality may apply ...

Give it a try! Often if you write something out, you can decide whether it is true. That works much better than sitting around waiting to be sure what you should do next.
Click to expand...
Wow I posted what you were thinking. You need to take some vitamins as I think you are slipping.
 
Yes, getting slow in my old age. That, and my time machine wasn't working, so I couldn't go back and post it before you as usual.
 
Dr.Peterson said:
Are you asking whether it is true that [MATH]|xy| = |x|\cdot |y|[/MATH]? or whether, if [MATH]|x|\le a[/MATH] and [MATH]|y|\le b[/MATH], then [MATH]|xy|\le ab[/MATH]?

You can certainly multiply things, and various properties of absolute value and of inequality may apply ...

Give it a try! Often if you write something out, you can decide whether it is true. That works much better than sitting around waiting to be sure what you should do next.
Click to expand...
1585430740740.pngI mean this... is multiplying those possible
or idk... i still don't understand what are you all trying to do....
 
So you are asking the second thing I suggested:
Dr.Peterson said:
[Is it true that], if [MATH]|x|\le a[/MATH] and [MATH]|y|\le b[/MATH], then [MATH]|xy|\le ab[/MATH]?
Click to expand...
You should be able to see this for yourself. First, let's make it match your question still more closely, while still keeping it general: ask yourself whether it is intuitively true that (assuming all numbers are non-negative):

If [MATH]a\le 1[/MATH] and [MATH]b\le c[/MATH], then [MATH]ab\le c[/MATH]​

Does it make sense, that is, that if you have two positive numbers b and c so that b is no larger than c, then if you multiply a by a positive number a that is less than 1, ab will still be no larger than c? The answer is, yes!

To make that intuition more formal, you can take it in steps:

[MATH]a\le 1[/MATH], so [MATH]a\cdot b\le 1\cdot b[/MATH] because you can multiply both sides of an inequality by a positive number, and it remains true; and since [MATH]b\le c[/MATH], you can conclude that [MATH]a\cdot b\le 1\cdot b = b\le c[/MATH], so [MATH]ab\le c[/MATH].​

So apply that to your problem!
 
Wernilou said:
Oh alright... it still kinda confuses me... but thanks.
Click to expand...
Dr.Peterson said:
You can lead a student to knowledge, but you can't make them think ...

I'll give it one more try.

You know that [MATH]|\sin(x) - \sin(y)| = 2\left|\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\right|[/MATH].

You know that [MATH]\left|\cos\left(\frac{x+y}{2}\right)\right| \le 1[/MATH].

You know that [MATH]\left|\sin\left(\frac{x-y}{2}\right)\right|\le \left|\frac{x-y}{2}\right|[/MATH].

Put those three things, all of which you yourself have said, together!
Click to expand...
Please look at Dr Peterson's post again. He gave you everything. Just put the pieces together.
 
I am done. Here is your answer. I hope that Dr Peterson does not mind.

[MATH]|\sin(x) - \sin(y)| = 2\left|\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\right|[/MATH] [Math]=2\left|\cos\left(\frac{x+y}{2}\right)\right|\left|\sin\left(\frac{x-y}{2}\right)\right|[/MATH][MATH]\leq 2\left|1\right|\left|\dfrac{x-y}{2}\right|=\left|x-y\right|[/MATH]
If you have any questions please ask.
 
You must log in or register to reply here.
Top