|sin a|<=|a|

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Luka 61727

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Can anyone prove it?

| sin a | <= | a |

I have a test tomorrow
 
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Luka 61727 said:
Can anyone prove it?

| sin a | <= | a |

I have a test tomorrow
Click to expand...
How to do it depends on your context.

What subject is the test on? In particular, what methods or concepts have you learned that you might use to prove this? What definition are you using for sin(x)? Are there any relevant theorems?

As an example, if the sine is defined geometrically in a unit circle, for acute angles, you might use the fact that any chord is shorter than the arc it subtends. But that may be totally irrelevant to your context.
 
In fact, I have no test, but my professor in the verbal command did this task but it was already the end of the day and I have to do tomorrow
 
Please provide some contextual information. If not a test (why did you say it was?), what is the course about? What definitions and theorems have you learned? What was the last thing the professor did before this?
 
I said that I have a test to somehow shortened the story. From the trigonometry we learned the additions formulas, equations, inequalities, the sinus and cosine theorem.
 
I need an answer urgently, I will depend on this, and I have only 9 hours to the beginning of the day.Now Is 22:46
 
Are you sitting in the exam as we speak?

Without SUBSTANTIAL input from you, that's just not what we do.

What are your thoughts? How might you proceed?
 
Luka 61727 said:
I said that I have a test to somehow shortened the story. From the trigonometry we learned the additions formulas, equations, inequalities, the sinus and cosine theorem.
Click to expand...
Here is what I asked for:
Dr.Peterson said:
What subject is the test on? In particular, what methods or concepts have you learned that you might use to prove this? What definition are you using for sin(x)? Are there any relevant theorems?
Click to expand...
I want to know what your definition of sine is. You probably are not aware that trigonometry can be taught in different ways, so you don't understand the importance of this. A proof must start from definitions and theorems; and in order to suggest a way, I need to actually see the definitions (and perhaps theorems) that you have available to use, not just to know that you have learned a definition.

Now, as I understand it, this is not for an exam, but for a very important assignment that you do not want to miss. I don't think you are asking us to do anything wrong. But if I were helping you face to face, I would be looking through your book or notes to find specific things on which you could base your proof; then I would point them out, and ask you to show me what you could do with them.
 
Luka 61727 said:
Can anyone prove it?
| sin a | <= | a |
Click to expand...
I was delighted when I saw the title of this post. But then I saw it was posted in the geometry forum not calculus. It is a wonderful application question. So I have decided to post the calculus solution even if it may not be helpful for your test.

If \(\displaystyle f(x)=\sin(x)~\&~a>0\) then the function \(\displaystyle f\) is continuous on \(\displaystyle [0,a]\) and differentiable on \(\displaystyle (0,a)\).
Let apply the mean value theorem. \(\displaystyle \exists c\in (0,a),\)
\(\displaystyle \begin{align*}\frac{f(a)-f(0)}{a-0}&=f'(c) \\\frac{\sin(a)-sin(0)}{a-0}&=\cos(c)\\\left|\frac{\sin(a)}{a}\right|&=|\cos(c)|\le 1\\\therefore~|\sin(a)|&\le |a|\end{align*}\)
 
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