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I need to show that $\sin(\frac{x}{n}) \frac{n}{x} \leq 1$. I know that this is equivalent as showing that $\frac{\sin(a)}{a} \leq 1$ with $a = \frac{x}{n}$, however I don't know how to show this. If $|a| > 1$ this is easy, and I know the limit when $a$ tends to $0$, but how can I show that this is true for all $a$ with absolute value between $0$ and $1$?

Gary
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matyce
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  • I suggest you will take a look at the pictures in this post: https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 , are they answering your question? – Udi Fogiel Mar 18 '22 at 23:28
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    Hint: $\frac{{\sin x}}{x} = \int_0^1 {\cos (xt)dt} $. – Gary Mar 18 '22 at 23:29
  • Welcome to MSE. Avoid the use of $*$ to denote multiplication, that's a practice in programming, but this is a math forum. Use a \cdot b to get $a \cdot b$, a \times b to get $a \times b$ or simple juxtaposition, as you did in the body of the question. – jjagmath Mar 18 '22 at 23:31
  • @Gary Since $|\cos(xt)| \leq 1$ then $\int_0^1 \cos(xt) dt \leq \int_0^1 1dt = 1$ ?Anyway thanks for your comment ! Sorry, I didn't know that I could use $\LaTeX$ in the title. – matyce Mar 18 '22 at 23:39
  • @matyce Yes, that is correct, but put absolute value around the cosine inside the integral. – Gary Mar 18 '22 at 23:41
  • @Gary If I put the absolute value around the cosine then can I say that $|\frac{\sin x}{x}| =| \int_0^1 \cos (xt) dt | \leq \int_0^1 | \cos (xt)| dt \leq 1$ ? – matyce Mar 18 '22 at 23:49
  • @matyce Yes, that is the correct way to do it. – Gary Mar 18 '22 at 23:52
  • See also https://math.stackexchange.com/a/3780744/269050. – Leandro Caniglia Mar 19 '22 at 11:17

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The area of the triangle is $\frac 12 |\sin a|$ The area of the section of the circle is $\frac 12 |a|.$

$|\sin a| \le |a|$

user317176
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